144.二叉树的前序遍历
前序遍历是:根-左-右
所以记录序列的的时候放在最前面
递归
class Solution {List<Integer> ans = new ArrayList<>();public List<Integer> preorderTraversal(TreeNode root) {if(root == null) return ans;ans.add(root.val);preorderTraversal(root.left);preorderTraversal(root.right);return ans;}
}
迭代
用栈模拟递归
- 首先访问根节点,存储根节点的值
- 随后访问左节点,每个左节点又是当前根节点,所以存储当前根节点的值,直到没有左节点就跳出循环
- 跳出循环后,访问当前节点的右节点,从栈顶取出当前节点,同时由于当前节点的左右节点都被要访问了,所以直接弹出当前节点,然后访问当前节点的右节点
- 右节点变成了当前根节点,记录值。回到步骤2判断是否有左节点
class Solution {public List<Integer> preorderTraversal(TreeNode root) {List<Integer> ans = new ArrayList<>();if(root == null) return ans;Deque<TreeNode> stack = new LinkedList<>();TreeNode node = root;while(node != null || !stack.isEmpty()){while(node != null){ans.add(node.val);stack.push(node);node = node.left;}node = stack.pop();node = node.right;}return ans;}
}
94.二叉树的中序遍历
递归
class Solution {List<Integer> ans = new ArrayList<>();public List<Integer> inorderTraversal(TreeNode root) {if(root == null) return ans;inorderTraversal(root.left);ans.add(root.val);inorderTraversal(root.right);return ans;}
}
迭代
class Solution {public List<Integer> inorderTraversal(TreeNode root) {List<Integer> ans = new ArrayList<>();if(root == null) return ans;Deque<TreeNode> stack = new LinkedList<>();TreeNode node = root;while(node != null || !stack.isEmpty()){while(node != null){stack.push(node);node = node.left;}node = stack.pop();ans.add(node.val);node = node.right;}return ans;}
}
145.二叉树的后序遍历
递归
class Solution {List<Integer> ans = new ArrayList<>();public List<Integer> postorderTraversal(TreeNode root) {if(root == null) return ans;postorderTraversal(root.left);postorderTraversal(root.right);ans.add(root.val);return ans;}
}
迭代
后序遍历:左-右-中
- 遍历完左子树后,需要从栈中获得当前节点,来获得右子树,但是后续还需要遍历中间节点,所以需要加回去栈中
- 但是当左右子树遍历完,就不需要再加回栈中了,只需要记录中间节点的值了
- 通过prev记录上一个遍历的右节点,然后判断当前节点的右节点和prev是否指向同一个,就可以判断当前节点是否还需要继续遍历右节点了
class Solution {public List<Integer> postorderTraversal(TreeNode root) {List<Integer> ans = new ArrayList<>();if(root == null) return ans;Deque<TreeNode> stack = new LinkedList<>();TreeNode node = root, prev = null;while(node != null || !stack.isEmpty()){while(node != null){stack.push(node);node = node.left;}node = stack.pop();if(node.right == null || prev == node.right){ans.add(node.val);prev = node;node = null;}else{stack.push(node);node = node.right;}}return ans;}
}
102.二叉树的层序遍历
BFS
class Solution {public List<List<Integer>> levelOrder(TreeNode root) {List<List<Integer>> ans = new ArrayList<>();if(root == null) return ans;Deque<TreeNode> queue = new LinkedList<>();queue.offer(root);while(!queue.isEmpty()){int curLevel = queue.size();List<Integer> ansLevel = new ArrayList<>();for(int i = 0; i < curLevel; ++ i){TreeNode node = queue.poll();ansLevel.add(node.val);if(node.left != null) queue.offer(node.left);if(node.right != null) queue.offer(node.right);}ans.add(ansLevel);}return ans;}
}
我的代码
class Solution {public List<List<Integer>> levelOrder(TreeNode root) {List<List<Integer>> ans = new ArrayList<>();if(root == null) return ans;List<TreeNode> levelNode = new ArrayList<>();levelNode.add(root);while(levelNode.size() != 0){List<Integer> levelAns = new ArrayList<>();List<TreeNode> nextLevelNode = new ArrayList<>();for(TreeNode node : levelNode){levelAns.add(node.val);System.out.println(node.val);if(node.left != null)nextLevelNode.add(node.left);if(node.right != null)nextLevelNode.add(node.right);}levelNode = nextLevelNode;ans.add(levelAns);}return ans;}
}
226.翻转二叉树
递归
class Solution {public TreeNode invertTree(TreeNode root) {if(root == null) return root;TreeNode right = invertTree(root.left);TreeNode left = invertTree(root.right);root.right = right;root.left = left;return root;}
}
101.对称二叉树
根节点的左右子树是对称的,所以让两个子树同时往下遍历
左子树的左节点等于右子树的右节点,反之亦然
并且如果是迭代则让应该相等的节点放在临近的一起
递归
class Solution {public boolean isSymmetric(TreeNode root) {return check(root.left, root.right);}public boolean check(TreeNode l, TreeNode r){if(l == null && r == null) return true;if(l == null || r == null) return false;boolean check1 = check(l.left, r.right);boolean check2 = check(l.right, r.left);return l.val == r.val && check1 && check2;}
}
迭代
注意遍历到叶子节点的时候要 continue
class Solution {public boolean isSymmetric(TreeNode root) {Deque<TreeNode> queue = new LinkedList<>();queue.offer(root.left);queue.offer(root.right);while(!queue.isEmpty()){TreeNode l = queue.poll();TreeNode r = queue.poll();if(l == null && r == null) continue;if(l == null || r == null || l.val != r.val) return false;queue.offer(l.left);queue.offer(r.right);queue.offer(l.right);queue.offer(r.left);}return true;}
}
104.二叉树的最大深度
递归写法:
class Solution {public int maxDepth(TreeNode root) {if(root == null) return 0;int lnum = maxDepth(root.left);int rnum = maxDepth(root.right);return Math.max(lnum, rnum) + 1;}
}
111.二叉树的最小深度
class Solution {public int minDepth(TreeNode root) {if(root == null) return 0;int lmin = minDepth(root.left);int rmin = minDepth(root.right);if(lmin == 0 && rmin == 0) return 1;else if(lmin == 0 || rmin == 0) return Math.max(lmin, rmin)+1;return Math.min(lmin, rmin)+1;}
}
222.完全二叉树的节点个数
完全二叉树的性质
分左右子树遍历:
- 如果左子树高度和右子树高度相等,则说明左子树已满,所以左子树的节点个数可以计算为 2 l L e v e l 2^{lLevel} 2lLevel,只需要计算右子树节点个数了
- 如果左子树高度和右子树高度不相等,则右子树已满,所以右子树的节点个数可以计算为 2 r L e v e l 2^{rLevel} 2rLevel,只需要计算左子树节点个数了
class Solution {public int countNodes(TreeNode root) {if(root == null) return 0;int lLevel = countLevel(root.left);int rLevel = countLevel(root.right);if(lLevel == rLevel){return countNodes(root.right) + (1<<lLevel);}else{return countNodes(root.left) + (1<<rLevel);}}public int countLevel(TreeNode root){int level = 0;while(root != null){++level;root = root.left;}return level;}
}
常规递归遍历
class Solution {public int countNodes(TreeNode root) {if(root == null) return 0;return countNodes(root.left) + countNodes(root.right) + 1;}
}
110.平衡二叉树
自顶向下递归
class Solution {public boolean isBalanced(TreeNode root) {if(root == null) return true;int lLevel = countLevel(root.left);int rLevel = countLevel(root.right);if(Math.abs(lLevel-rLevel) > 1) return false;return isBalanced(root.left) && isBalanced(root.right);}public int countLevel(TreeNode root){if(root == null) return 0;return Math.max(countLevel(root.left), countLevel(root.right)) + 1;}
}
countLevel被重复调用了
自底向上递归
- 如果平衡就返回高度
- 如果不平衡就返回一个负数
- 后面判断是否子树已经不平衡,通过高度是否是负数检测
class Solution {public boolean isBalanced(TreeNode root) {return countLevel(root) >= 0;}public int countLevel(TreeNode root){if(root == null) return 0;int lLevel = countLevel(root.left);int rLevel = countLevel(root.right);if(lLevel == -1 || rLevel == -1 || Math.abs(lLevel-rLevel) > 1){return -1;}return Math.max(lLevel, rLevel) + 1;}
}
左子树已经不平衡,右子树就没必要递归下去了,直接 return
class Solution {public boolean isBalanced(TreeNode root) {return countLevel(root) >= 0;}public int countLevel(TreeNode root){if(root == null) return 0;int lLevel = countLevel(root.left);if(lLevel == -1) return -1;int rLevel = countLevel(root.right);if(rLevel == -1) return -1;if(Math.abs(lLevel-rLevel) > 1){return -1;}return Math.max(lLevel, rLevel) + 1;}
}
257.二叉树的所有路径
遇到叶子节点就把路径计入答案,否则把当前节点计入路径中
class Solution {public List<String> binaryTreePaths(TreeNode root) {List<String> paths = new ArrayList<>();TreePath(root, "", paths);return paths;}public void TreePath(TreeNode root, String path, List<String> paths){if(root == null) return;if(root.left == null && root.right == null){paths.add(path + root.val);return;}path += root.val + "->";TreePath(root.left, path, paths);TreePath(root.right, path, paths);}
}
String 的加法会不断的创建拷贝生成新的字符串,耗费时间和空间。可以用 StringBuilder 或者 StringBuffer 优化
但是它们属于引用类型,后面对其字符串的更改,回溯后更改依然存在,所以在回溯后需要delete掉多余的内容
class Solution {public List<String> binaryTreePaths(TreeNode root) {List<String> paths = new ArrayList<>();TreePath(root, new StringBuilder(), paths);return paths;}public void TreePath(TreeNode root, StringBuilder path, List<String> paths){if(root == null) return;if(root.left == null && root.right == null){paths.add(path.toString() + root.val); // toString 一定在前,+val在后return;}int prevLen = path.length();path.append(root.val).append("->");TreePath(root.left, path, paths);TreePath(root.right, path, paths);path.setLength(prevLen);}
}
404.左叶子之和
https://leetcode.cn/problems/sum-of-left-leaves/
- 首先必须有左节点
- 然后必须是叶子节点
- 每次的递归累计值不能丢失
class Solution {public int sum = 0;public int sumOfLeftLeaves(TreeNode root) {if(root == null) return sum;if(root.left != null && root.left.left == null && root.left.right == null)sum = sum + root.left.val;sumOfLeftLeaves(root.left);sumOfLeftLeaves(root.right);return sum;}
}
513.找树左下角的值
左下角的值,也就是从左往右dfs最深的一层的第一个节点
或者,bfs从左往右的第一个节点(从右往左的最后一个节点)
dfs
记录选择节点的深度
public class Solution {int curDepth = 1;int maxDepth = 0;int curVal = 0;public int findBottomLeftValue(TreeNode root) {if(root == null) return curVal;if(root.left == null && root.right == null){if(maxDepth < curDepth){maxDepth = curDepth;curVal = root.val;return curVal;}}curDepth += 1;findBottomLeftValue(root.left);findBottomLeftValue(root.right);curDepth -= 1;return curVal;}
}
112.路径总和
easy
求二叉树路径上所有节点的和是否等于某个数
每次向下递归的时候减去当前节点的值,直到叶子节点看是否减到0了
左右子树只要有一个返回 true,就存在
class Solution {public boolean hasPathSum(TreeNode root, int targetSum) {if(root == null) return false;if(root.left == null && root.right == null){if(targetSum - root.val == 0) return true;return false;}return hasPathSum(root.left, targetSum - root.val) || hasPathSum(root.right, targetSum - root.val);}
}
106.从中序与后序遍历序列构造二叉树
中序遍历:左 [中] 右 :获取左右子树的范围
后序遍历:左 右 [中] :获取根节点
所以后序遍历的最后一个节点一定是根节点
- 首先根据后序遍历,其最后一个位置就是根节点idx
- 找到根节点idx对应中中序遍历的位置 mid ,获取左右子树的范围 [l, mid-1], [mid+1, r]
- 递归的在 [l, mid-1], [mid+1, r] 区间重复上述两个过程;
- 由于后序遍历是先左,再右,所以下一个根节点也就是–idx是右子树的根节点。所以必须先递归右区间,再递归左区间
class Solution {Map<Integer, Integer> hash;int idx;public TreeNode buildTree(int[] inorder, int[] postorder) {hash = new HashMap<>();for(int i = 0; i < inorder.length; ++ i){hash.put(inorder[i], i);}idx = inorder.length-1;return build(inorder, postorder, 0, inorder.length-1);}public TreeNode build(int[] inorder, int[] postorder, int l, int r){if(l > r){return null;}// if(l == r){// idx--;// return new TreeNode(inorder[l]);// }TreeNode root = new TreeNode(postorder[idx]);int mid = hash.get(postorder[idx--]);root.right = build(inorder, postorder, mid + 1, r);root.left = build(inorder, postorder, l, mid-1);return root;}
}
654.最大二叉树
给定一个不重复的整数数组 nums 。 最大二叉树 可以用下面的算法从 nums 递归地构建:
- 创建一个根节点,其值为 nums 中的最大值。
- 递归地在最大值 左边 的 子数组前缀上 构建左子树。
- 递归地在最大值 右边 的 子数组后缀上 构建右子树。
递归
每次在指定区间内找到最大值作为根节点
class Solution {public TreeNode constructMaximumBinaryTree(int[] nums) {return build(nums, 0, nums.length-1);}public TreeNode build(int[] nums, int l, int r){if(l > r){return null;}int mid = l;for(int i = l + 1; i <= r; ++ i){if(nums[i] > nums[mid]){mid = i;}}TreeNode root = new TreeNode(nums[mid]);root.left = build(nums, l, mid - 1);root.right = build(nums, mid + 1, r);return root;}
}
单调栈
找到每个节点,左边第一个和右边第一个比它大的元素,其中较小的元素为当前节点的父节点
找到每个元素第一个比它大的元素可以用单调栈:
构造一个单调递减的栈,每当加入当前元素:
- 如果栈顶元素比当前元素小,那么当前元素就是栈顶元素右边第一个比它大的
- 如果栈顶元素比当前元素大,那么栈顶元素就是当前元素左边第一个比它大的
class Solution {public TreeNode constructMaximumBinaryTree(int[] nums) {int n = nums.length;int[] left = new int[n];int[] right = new int[n];Arrays.fill(left, -1);Arrays.fill(right, -1);Deque<Integer> stack = new ArrayDeque<>();TreeNode[] nodes = new TreeNode[n];for(int i = 0; i < n; ++ i){nodes[i] = new TreeNode(nums[i]);while(!stack.isEmpty() && nums[i] > nums[stack.peek()]){right[stack.pop()] = i;}if(!stack.isEmpty()){left[i] = stack.peek();}stack.push(i);}TreeNode root = null;for(int i = 0; i < n; ++ i){if(left[i] == -1 && right[i] == -1){root = nodes[i];}else if(left[i] == -1 || (right[i] != -1 && nums[left[i]] > nums[right[i]])){nodes[right[i]].left = nodes[i];}else{nodes[left[i]].right = nodes[i];}}return root;}
}
617.合并二叉树
class Solution {public TreeNode mergeTrees(TreeNode root1, TreeNode root2) {if(root1 == null && root2 == null){return null;}int val1 = root1 == null ? 0 : root1.val;int val2 = root2 == null ? 0 : root2.val;TreeNode root = new TreeNode(val1 + val2);root.left = mergeTrees(root1 == null ? null : root1.left, root2 == null ? null : root2.left);root.right = mergeTrees(root1 == null ? null : root1.right, root2 == null ? null : root2.right);return root;}
}
更简介的代码:
不需要判断两个通知为 null 才返回 null
只要有一方为 null,就返回另一方
class Solution {public TreeNode mergeTrees(TreeNode root1, TreeNode root2) {if(root1 == null){return root2;}if(root2 == null){return root1;}TreeNode root = new TreeNode(root1.val + root2.val);root.left = mergeTrees(root1.left, root2.left);root.right = mergeTrees(root1.right, root2.right);return root;}
}
700.二叉搜索树中的搜索
class Solution {public TreeNode searchBST(TreeNode root, int val) {if(root == null){return null;}if(root.val == val){return root;}if(root.val > val){return searchBST(root.left, val);}return searchBST(root.right, val);}
}
98.验证二叉搜索树
递归判断区间范围
class Solution {public boolean isValidBST(TreeNode root) {return check(root, Long.MIN_VALUE, Long.MAX_VALUE);}public boolean check(TreeNode root, long left, long right){if(root == null) return true;if(root.val <= left || root.val >= right) return false;return check(root.left, left, root.val) && check(root.right, root.val, right);}
}
中序遍历
迭代版本的,便于拿到前一个节点的值
530.二叉搜索树的最小绝对差
二叉搜索树的中序遍历结果中,相邻数字相减的最小值
迭代:
class Solution {public int getMinimumDifference(TreeNode root) {Deque<TreeNode> stack = new ArrayDeque<>();int pre = -1, res = Integer.MAX_VALUE;while(root != null || !stack.isEmpty()){while(root != null){stack.push(root);root = root.left;}root = stack.pop();if(pre != -1){res = Math.min(res, root.val - pre);}pre = root.val;root = root.right;}return res;}
}
中序遍历模板:
class Solution{public void minSearch(TreeNode root){Deque<TreeNode> stack = new ArrayDeque<>();// 节点不为空,或者栈里面还有剩下没被遍历的左节点们while(root != null || !stack.isEmpty()){// 中序遍历,先遍历到极左:while(root != null){stack.push(root);root = root.left;}// 中序遍历,极左遍历完了,遍历中间的节点:root = stack.pop(); // 栈中弹出最后遍历到的左节点root = root.rigth; // 最后遍历右节点}}
}
递归:
class Solution {int pre = -1;int ans = Integer.MAX_VALUE;public int getMinimumDifference(TreeNode root) {if(root == null){return ans;}getMinimumDifference(root.left);if(pre != -1){ans = Math.min(ans, root.val - pre);}pre = root.val;getMinimumDifference(root.right);return ans;}
}
236.二叉树的最近公共祖先
dfs
回溯的时候判断,该节点 x 本身及其子节点是否包含p或q中的一个,判断条件:
-
x == p || x == q || lson || rson
-
如果一个节点 x 的左子树 lson 和 右子树 rson 各自包含一个 p 或者 q 节点,那么 x 就是最近公共祖先
-
如果一个节点 x 本身是 p 或者 q 节点,x 的 lson 或者 rson 包含剩下的 p 或者 q 节点,那么 x 就是最近公共祖先
转换为:
- (lson && rson) || ((x == p || x == q) && (lson || rson))
class Solution {TreeNode ans;public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {ans = null;dfs(root, p, q);return ans;}public boolean dfs(TreeNode root, TreeNode p, TreeNode q){if(root == null) return false;boolean lson = dfs(root.left, p, q);boolean rson = dfs(root.right, p, q);if((lson && rson) || ((root.val == p.val || root.val == q.val)&&(lson || rson))){ans = root;return true;}return root.val == p.val || root.val == q.val || lson || rson;}
}
优雅的解法:
class Solution {public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {if (root == null || root == p || root == q) {return root; // 1)一个是另外一个的祖先}TreeNode left = lowestCommonAncestor(root.left, p, q);TreeNode right = lowestCommonAncestor(root.right, p, q);if (left != null && right != null) {return root; // 2)左右两边各自有一个o1、o2,返回这个祖先"}return left != null ? left : right;// 1) / 2) 找不到,回溯时一直是null,如果找到了,那么将找到的值往上窜!}
}
存储父节点
从根节点 dfs,存储每个节点的父节点
从 p 开始向上搜索,将遇到的节点进行记录
再次从 q 开始向上搜索,如果遇到了之前遇到的节点,这个节点就是最近公共祖先
235.二叉搜索树的最近公共祖先
一次搜索两个节点
如果 p.val <= root.val <= q.val 那么 root 就是 LCA
class Solution {TreeNode ans = null;public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {ans = null;dfs(root, p, q);return ans;}public void dfs(TreeNode root, TreeNode p, TreeNode q){if(root == null) return;if(root.val > p.val && root.val > q.val) dfs(root.left, p, q);else if(root.val < p.val && root.val < q.val) dfs(root.right, p, q);else ans = root;}
}
两次搜索
两次搜索获取两个目标节点的路径,路径的分叉点就是 LCA
class Solution {public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {List<TreeNode> path_p = new ArrayList<>();List<TreeNode> path_q = new ArrayList<>();dfs(path_p, root, p);dfs(path_q, root, q);int i = 0;TreeNode ans = null;while(i < path_p.size() && i < path_q.size()){if(path_p.get(i).val == path_q.get(i).val){ans = path_p.get(i);}else break;++i;}return ans;}public void dfs(List<TreeNode> path, TreeNode root, TreeNode target){path.add(root);if(root == target) return;if(root.val > target.val) dfs(path, root.left, target);else dfs(path, root.right, target);}
}
701.二叉搜索树中的插入操作
优雅写法
用递归时的方向直接判断左还是右
返回 root 节点给父节点,重建树
class Solution {boolean add = false;public TreeNode insertIntoBST(TreeNode root, int val) {return dfs(root, val);}public TreeNode dfs(TreeNode root, int val){if(root == null) return new TreeNode(val);if(val < root.val){root.left = dfs(root.left, val);}else{root.right = dfs(root.right, val);}return root;}
}
我的代码
找到插入的叶子节点,判断插入的方向。插入后限制不可插入:
class Solution {boolean add = false;TreeNode node = null;public TreeNode insertIntoBST(TreeNode root, int val) {node = new TreeNode(val);dfs(root, val);return root == null ? node : root;}public void dfs(TreeNode root, int val){if(root == null) return;int dir = 0;if(root.val > val) {dfs(root.left, val);dir = 1;}else dfs(root.right, val);if(add == false){if(dir == 0){root.right = node;}else{root.left = node;}add = true;}}
}
450.删除二叉搜索树中的节点
首先找到要删除的节点,然后分类讨论:
- 如果删除节点左右子树为空,则直接返回 null
- 如果删除节点左子树为空,返回右子树
- 如果删除节点右子树为空,返回左子树
- 如果删除节点左右子树都不为空:
- 找到删除节点的后继节点,也就是 delete 节点的右子树的最左节点 node
- 该 node 节点是仅仅只大于 delete节点的,所以用 node 代替 delete 节点,二叉树的性质保持不变
- 替代方法:首先删除 node 节点,然后改变 node 的左右子树
class Solution {public TreeNode deleteNode(TreeNode root, int key) {if(root == null) return root;if(key < root.val){root.left = deleteNode(root.left, key);return root;}else if(key > root.val){root.right = deleteNode(root.right, key);return root;}else if(key == root.val){if(root.left == null && root.right == null){return null;}else if(root.left == null){return root.right;}else if(root.right == null){return root.left;}else{TreeNode node = root.right;while(node.left != null){node = node.left;}// 这里要写 root.right = deleteNode(xx),因为有可能删除的是root.right,就会返回 nullroot.right = deleteNode(root.right, node.val);node.left = root.left;node.right = root.right;return node;}}return root;}
}
696.修剪二叉搜索树
将不在区间 [low, high] 内的节点剪掉
-
如果节点的值 小于 low ,则只用修剪右子树
-
如果节点的值 大于 high,则只用修建左子树
-
如果位于区间,则其左右子树可能不正常,所以递归修建左右子树
-
修剪到最后,如果没有正常的节点则会一直递归到节点为 null,返回 null
-
如果遇到正常节点,则返回该节点本身
class Solution {public TreeNode trimBST(TreeNode root, int low, int high) {if(root == null) return root;if(root.val < low){return trimBST(root.right, low, high);}else if(root.val > high){return trimBST(root.left, low, high);}else{root.left = trimBST(root.left, low, high);root.right = trimBST(root.right, low, high);return root;}}
}
108.将有序数组转换为二叉搜索树
每次选择区间内的中间值作为根节点,递归构造左右子树,
直到区间唯一个点返回该点构造的节点,或者 l > r 说明不存在左边或者右边的值,返回 null
返回左右子树构造好的根节点给上层的根节点构造左右子树
class Solution {public TreeNode sortedArrayToBST(int[] nums) {return build(nums, 0, nums.length - 1);}public TreeNode build(int[] nums, int l, int r){if(l == r){return new TreeNode(nums[l]);}else if(l > r){return null;}int mid = l + (r - l)/2;TreeNode root = new TreeNode(nums[mid]);root.left = build(nums, l, mid - 1);root.right = build(nums, mid + 1, r);return root;}
}
538.把二叉搜索树转换为累加树
反向中序遍历,累加之前的值:
class Solution {int sum = 0;public TreeNode convertBST(TreeNode root) {if(root == null) return root;convertBST(root.right);sum += root.val;root.val = sum;convertBST(root.left);return root;}
}
我的代码:
先dfs一遍算出总和
中序遍历一遍,减去之前的值
class Solution {int sum = 0;public TreeNode convertBST(TreeNode root) {dfs(root);if(root != null) sum += root.val;build(root);return root;}public int dfs(TreeNode root){if(root == null) return 0;int l = dfs(root.left);int r = dfs(root.right);sum = sum + l + r;return root.val;}public void build(TreeNode root){if(root == null) return;build(root.left);int sub = root.val;root.val = sum;sum -= sub;build(root.right);}
}