递归遍历基础篇
前序遍历,中序遍历,后序遍历是根据处理根节点的位置来命名的。
树的处理大多用到了递归,递归需要知道终止条件。
前序遍历(中左右)
144.二叉树的前序遍历
中左右,先处理根节点,再处理左子树,再处理右子树
class Solution {public List<Integer> preorderTraversal(TreeNode root) {List<Integer> res = new ArrayList<>();preOrder(res, root);return res;}private void preOrder(List<Integer> res, TreeNode root) {if(root==null){return;}res.add(root.val);preOrder(res, root.left);preOrder(res, root.right);}
}
非递归版实现前序遍历
使用栈,当前节点处理完,先塞入右节点(后处理),再塞入左节点。
class Solution {public List<Integer> preorderTraversal(TreeNode root) {Stack<TreeNode> stack = new Stack<>();List<Integer> res = new ArrayList<>();if (root == null) {return res;}//右左stack.add(root);while (!stack.isEmpty()) {TreeNode pop = stack.pop();res.add(pop.val);if (pop.right != null) {stack.push(pop.right);}if (pop.left != null) {stack.push(pop.left);}}return res;}
}
中序遍历(左中右)
94.二叉树的中序遍历
递归方式
class Solution {public List<Integer> inorderTraversal(TreeNode root) {List<Integer> res = new ArrayList<>();inOrder(res, root);return res;}private void inOrder(List<Integer> res, TreeNode root) {if (root == null) {return;}inOrder(res, root.left);res.add(root.val);inOrder(res, root.right);}
}
非递归,左中右,先找到最左节点,处理当前节点,处理右节点。把最左边的节点都压入栈中。
class Solution {public List<Integer> inorderTraversal(TreeNode root) {List<Integer> res = new ArrayList<>();if (root == null) {return res;}Stack<TreeNode> stack = new Stack<>();TreeNode cur = root;while (cur != null || !stack.isEmpty()) {if (cur != null) {stack.push(cur);cur = cur.left;} else {TreeNode pop = stack.pop();res.add(pop.val);cur = pop.right;}}return res;}
后序遍历(左右中)
145.二叉树的后序遍历
递归方式
class Solution {public List<Integer> postorderTraversal(TreeNode root) {List<Integer> res = new ArrayList<>();postOrder(res, root);return res;}public void postOrder(List<Integer> res, TreeNode root) {if (root == null) {return;}postOrder(res,root.left);postOrder(res,root.right);res.add(root.val);}
}
非递归方式,左右中的逆序是中右左,处理可以参考前序遍历,最后进行倒序。
class Solution {public List<Integer> postorderTraversal(TreeNode root) {List<Integer> res = new ArrayList<>();if (root == null) {return res;}Stack<TreeNode> stack = new Stack<>();stack.push(root);while (!stack.isEmpty()) {TreeNode pop = stack.pop();res.add(pop.val);if(pop.left!=null){stack.push(pop.left);}if(pop.right!=null){stack.push(pop.right);}}Collections.reverse(res);return res;}}
层序遍历
LeetCode199.二叉树的右视图
LeetCode199.二叉树的右视图
获取当前行的最后一个元素。
class Solution_LC199 {public List<Integer> rightSideView(TreeNode root) {List<Integer> res = new ArrayList<>();if (root == null) {return res;}Queue<TreeNode> queue = new LinkedList<>();queue.offer(root);while (!queue.isEmpty()) {int size = queue.size();while (size > 0) {size--;TreeNode node = queue.poll();if (size == 0) {res.add(node.val);}if (node.left != null) {queue.offer(node.left);}if (node.right != null) {queue.offer(node.right);}}}return res;}
}
LeetCode103.二叉树的锯齿形层序遍历
LeetCode103.二叉树的锯齿形层序遍历
对树进行判空
获取当前行的元素,需要获取队列的大小。
class Solution {public List<List<Integer>> zigzagLevelOrder(TreeNode root) {List<List<Integer>> res = new ArrayList<>();if (root == null) {return res;}Queue<TreeNode> queue = new LinkedList<>();queue.offer(root);int index = 0;while (!queue.isEmpty()) {index++;LinkedList<Integer> list = new LinkedList<>();int size = queue.size();while (size > 0) {size--;TreeNode node = queue.poll();if (index % 2 == 0) {list.addFirst(node.val);} else {list.addLast(node.val);}if (node.left != null) {queue.add(node.left);}if (node.right != null) {queue.add(node.right);}}res.add(list);}return res;}}
剑指 Offer II 044. 二叉树每层的最大值
剑指 Offer II 044. 二叉树每层的最大值
原理同上。
class Solution_JZ044 {public List<Integer> largestValues(TreeNode root) {List<Integer> res = new ArrayList<>();if (root == null) {return res;}Queue<TreeNode> queue = new LinkedList<>();queue.offer(root);while (!queue.isEmpty()) {int size = queue.size();int max = Integer.MIN_VALUE;while (size > 0) {size--;TreeNode node = queue.poll();if (node.val > max) {max = node.val;}if (node.left != null) {queue.offer(node.left);}if (node.right != null) {queue.offer(node.right);}}res.add(max);}return res;}
}
Leetcode101. 对称二叉树
Leetcode101. 对称二叉树
递归
比较最左边和最右边的元素。
class Solution {public boolean isSymmetric(TreeNode root) {if (root == null) {return true;}return dfs(root.left, root.right);}private boolean dfs(TreeNode left, TreeNode right) {if (left == null && right == null) {return true;}if (left == null || right == null) {return false;}if (left.val != right.val) {return false;}return dfs(left.left, right.right) && dfs(right.left, left.right);}
}
使用队列
class Solution {public boolean isSymmetric(TreeNode root) {if (root == null || (root.left == null && root.right == null)) {return true;}Queue<TreeNode> queue = new LinkedList<>();queue.offer(root.left);queue.offer(root.right);while (!queue.isEmpty()) {TreeNode left = queue.poll();TreeNode right = queue.poll();if (left == null && right == null) {continue;}if (left == null || right == null) {return false;}if (left.val != right.val) {return false;}queue.offer(left.left);queue.offer(right.right);queue.offer(right.left);queue.offer(left.right);}return true;}
}
前序遍历
LeetCode257.二叉树的所有路径
LeetCode257: 二叉树的所有路径
结束条件:该节点是叶子节点,结果集添加
path
如何添加连接字符
->
。
class Solution {List<String> res = new ArrayList<>();public List<String> binaryTreePaths(TreeNode root) {dfs(root, "");return res;}private void dfs(TreeNode root, String s) {if (root == null) {return;}s = s + root.val;if (root.left == null && root.right == null) {res.add(s);}else {dfs(root.left, s + "->");dfs(root.right, s + "->");}}
}
LeetCode129.求根到叶子节点数字之和
LeetCode 129.求根到叶子节点数字之和
按照上题的思路来,先把结果添加到集合里面。
class Solution {List<Integer> res = new ArrayList<>();public int sumNumbers(TreeNode root) {if (root == null) {return 0;}dfs(root, 0);int sum = 0;for (int i = 0; i < res.size(); i++) {sum += res.get(i);}return sum;}private void dfs(TreeNode root, int num) {if (root == null) {return;}num = num * 10 + root.val;if (root.left == null && root.right == null) {res.add(num);} else {dfs(root.left, num);dfs(root.right, num);}}
}
递归,当前节点的结果等于左子树操作+右子树操作
class Solution {List<Integer> res = new ArrayList<>();public int sumNumbers(TreeNode root) {if (root == null) {return 0;}return dfs2(root, 0);}private int dfs2(TreeNode root, int pre) {if (root == null) {return 0;}int sum = pre * 10 + root.val;if (root.left == null && root.right == null) {return sum;}return dfs2(root.right, sum) + dfs2(root.left, sum);}
}
LeetCode112.路径总和
112. 路径总和
不断累加值,判断是否和目标值相等
class Solution {public boolean hasPathSum(TreeNode root, int targetSum) {return dfs(root, targetSum, 0);}private boolean dfs(TreeNode root, int targetSum, int pre) {if (root == null) {return false;}if (root.left == null && root.right == null) {return pre + root.val == targetSum;} else {return dfs(root.left, targetSum, pre + root.val) || dfs(root.right, targetSum, pre + root.val);}}
}
更新目标值
class Solution {public boolean hasPathSum(TreeNode root, int targetSum) {if (root == null) {return false;}if (root.left == null && root.right == null) {return targetSum == root.val;}return hasPathSum(root.left, targetSum - root.val) || hasPathSum(root.right, targetSum - root.val);}
}
LeetCode113.路径总和II
113. 路径总和 II
用队列记录当前路径下的元素,方便回退。
class Solution {List<List<Integer>> res = new LinkedList<>();LinkedList<Integer> queue = new LinkedList<>();public List<List<Integer>> pathSum(TreeNode root, int targetSum) {dfs(root,targetSum );return res;}private void dfs(TreeNode root, int targetSum) {if (root == null) {return;}targetSum -= root.val;queue.offer(root.val);if (root.left == null && root.right == null && targetSum == 0) {res.add(new ArrayList<>(queue));} else {dfs(root.left, targetSum);dfs(root.right, targetSum);}queue.pollLast();}
}
LeetCode437.路径总和 III(**)
LeetCode437. 路径总和 III
定义一个以某节点为开始节点获取路径总和为
targetSum
的方法。不一定是叶子节点
用long代替int,因为有测试用例减溢出。
class Solution {public int pathSum(TreeNode root, int targetSum) {if (root == null) {return 0;}int res = rootSum(root, targetSum);return res + pathSum(root.left, targetSum) + pathSum(root.right, targetSum);}private int rootSum(TreeNode root, long targetSum) {int ret = 0;if (root == null) {return 0;}if (targetSum == root.val) {ret++;}return ret + rootSum(root.left, targetSum - root.val) + rootSum(root.right, targetSum - root.val);}
}
中序遍历
LeetCode98.验证二叉搜索树
LeetCode98.验证二叉搜索树
左子树上的节点都小于根节点,右子树上的值都大于根节点。
递归。
class Solution {public boolean isValidBST(TreeNode root) {return isValidBST(root, Long.MAX_VALUE, Long.MIN_VALUE);}private boolean isValidBST(TreeNode root, long maxValue, long minValue) {if (root == null) {return true;}if (root.val >= maxValue || root.val <= minValue) {return false;}return isValidBST(root.left, root.val, minValue) && isValidBST(root.right, maxValue, root.val);}
}
中序遍历
中序遍历 ,先左再中后右,左<中<右
比较数组是否有序,用tmp来缓存最小元素。
class Solution {public boolean isValidBST(TreeNode root) {Stack<TreeNode> stack = new Stack<>();long tmp = Long.MIN_VALUE;TreeNode cur = root;while (!stack.isEmpty() || cur != null) {if (cur != null) {stack.push(cur);cur = cur.left;}else {cur = stack.pop();if (cur.val <= tmp) {return false;}tmp = cur.val;cur = cur.right;}}return true;}
}
剑指 Offer 54. 二叉搜索树的第k大节点
剑指 Offer 54. 二叉搜索树的第k大节点.
第n大,就是数组中第n-k个元素。第2大的元素逆序在索引1的位置。
右中左,即中序遍历的数组的倒序。
class Solution_LCR174 {int res;int index = 0;public int findTargetNode(TreeNode root, int cnt) {dfs(root, cnt);return res;}private void dfs(TreeNode root, int cnt) {if (root == null) {return;}dfs(root.right, cnt);index++;if (index == cnt) {res = root.val;}dfs(root.left, cnt);}
}
Leetcode230. 二叉搜索树中第K小的元素
Leetcode230. 二叉搜索树中第K小的元素
同上,不需要倒序,更简单
class Solution {int k;int res;int index = 0;public int kthSmallest(TreeNode root, int k) {this.k = k;dfs(root);return res;}private void dfs(TreeNode root) {if (root == null) {return;}dfs(root.left);index++;if (index == k) {res = root.val;}dfs(root.right);}
}
剑指offer36: 二叉搜索树与双向链表(*)
剑指offer36: 二叉搜索树与双向链表
pre节点是前节点,获取最左边的节点为pre节点。不断更新pre节点。
核心还是中序遍历。
class Solution {Node pre = null, head = null;public Node treeToDoublyList(Node root) {if (root == null) return root;dfs(root);head.left = pre;pre.right = head;return head;}void dfs(Node root) {if (root == null) {return;}dfs(root.left);if (pre != null) {pre.right = root;} else {head = root;}root.left = pre;pre = root;dfs(root.right);}
}
Leetcode.538. 把二叉搜索树转换为累加树(*)
Leetcode.538. 把二叉搜索树转换为累加树.
理解题意,原先是二叉搜索树,当先不一定是二叉搜索树。当前节点为原有树的大于等于当前节点(当前节点和所有右子树节点)的和。
class Solution {int sum = 0;public TreeNode convertBST(TreeNode root) {if (root == null) {return null;}convertBST(root.right);sum += root.val;root.val = sum;convertBST(root.left);return root;}
}
后序遍历
Leetcode104. 二叉树的最大深度
Leetcode104. 二叉树的最大深度
方法一:递归(后续遍历,左右中)
方法二:层序遍历
class Solution {public int maxDepth(TreeNode root) {if (root == null) {return 0;} else {return Math.max(maxDepth(root.right), maxDepth(root.left)) + 1;}}
}
LeetCode226. 翻转二叉树
LeetCode226. 翻转二叉树
要暂存左边的树
class Solution {public TreeNode invertTree(TreeNode root) {if (root == null) {return null;}TreeNode left = invertTree(root.left);TreeNode right = invertTree(root.right);root.right = left;root.left = right;return root;}
}
Leetcode110. 判断是否是平衡二叉树
Leetcode110. 判断是否平衡二叉树
求深度的扩展
class Solution {public boolean isBalanced(TreeNode root) {if (root == null) {return true;}return Math.abs(getDepth(root.left) - getDepth(root.right)) <= 1 & isBalanced(root.left)&& isBalanced(root.right);}private int getDepth(TreeNode root) {if (root == null) {return 0;}return 1 + Math.max(getDepth(root.right), getDepth(root.left));}
}
LeetCode543.二叉树的直径
LeetCode543.二叉树的直径
结果最大的数据不一定是根节点。
每一个节点的对应的直径等于左子树的深度+右子树的深度。
class Solution {int ans;public int diameterOfBinaryTree(TreeNode root) {if (root == null) {return 0;}depth(root);return ans;}private int depth(TreeNode root) {if (root == null) {return 0;}int left = depth(root.left);int right = depth(root.right);ans = Math.max(ans, right + left);return 1 + Math.max(left, right);}
}
LeetCode124.二叉树中的最大路径和
LeetCode124.二叉树中的最大路径和
计算以某节点为起点,最长的路径。路径和=左最大路径+右最大路径+当前节点的值。
如果该路径大于0,则选;否则不选。
class Solution {int ans = Integer.MIN_VALUE;public int maxPathSum(TreeNode root) {maxGain(root);return ans;}public int maxGain(TreeNode node) {if (node == null) {return 0;}int left = Math.max(maxGain(node.left), 0);int right = Math.max(maxGain(node.right), 0);ans = Math.max(ans, node.val + left + right);return node.val + Math.max(left, right);}
}
Leetcode236. 二叉树的最近公共祖先(**)
Leetcode236. 二叉树的最近公共祖先
左子树能找到,两个节点都在左边;左右子树都找到,返回root;左右子树都找不到,返回
class Solution {public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {if (root == null || p == root || q == root) {return root;}TreeNode leftNode = lowestCommonAncestor(root.left, p, q);TreeNode rightNode = lowestCommonAncestor(root.right, p, q);if (leftNode == null && rightNode == null) {return null;}if (leftNode == null) {return rightNode;}if (rightNode == null) {return leftNode;}return root;}
}
二叉搜索树
Leetcode235. 二叉搜索树的最近公共祖先
Leetcode235. 二叉搜索树的最近公共祖先
遍历树的元素
class Solution {public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {while (true) {if (root.val > p.val && root.val > q.val) {root = root.left;} else if (root.val < p.val && root.val < q.val) {root = root.right;} else {break;}}return root;}
}
剑指offer33: 验证二叉搜索树的后序遍历序列
验证二叉搜索树的后序遍历序列
后续遍历是左右中的结构,最后一个元素是root
序列中第一个元素大于root元素,该元素及以后是右子树的序列。
最后一个元素是根元素,不是右子树的一部分。
class Solution {public boolean verifyTreeOrder(int[] postorder) {// 左右中return dfs(postorder, 0, postorder.length - 1);}private boolean dfs(int[] postorder, int left, int right) {if (left >= right) {return true;}int tmp = left;int root = postorder[right];while (tmp < right && postorder[tmp] < root) {tmp++;}int mid = tmp;while (tmp < right) {if (postorder[tmp] < root) {return false;}tmp++;}return dfs(postorder, left, mid - 1) && dfs(postorder, mid, right - 1);}
}
二叉树的修改构造
LeetCode105.从前序和中序遍历构造二叉树
LeetCode105.从前序和中序遍历构造二叉树
前序 中左右 中序 左中右
分割中序遍历的序列,获取到左子树的长度和右子树的长度。
对数组的操作,要么是拷贝数组,要么是指定索引。
class Solution {Map<Integer, Integer> map = new HashMap<>();public TreeNode buildTree(int[] preorder, int[] inorder) {for (int i = 0; i < inorder.length; i++) {map.put(inorder[i], i);}return buildTree(preorder, 0, preorder.length - 1, inorder, 0, inorder.length - 1);}private TreeNode buildTree(int[] preorder, int preLeft, int preRight, int[] inorder, int inLeft, int inRight) {if (preRight < preLeft || inRight < inLeft) {return null;}int rootIndex = map.get(preorder[preLeft]);TreeNode root = new TreeNode(preorder[preLeft]);int leftLength = rootIndex - inLeft;root.left = buildTree(preorder, preLeft + 1, preLeft + leftLength, inorder, inLeft, rootIndex);root.right = buildTree(preorder, preLeft + leftLength + 1, preRight, inorder, rootIndex + 1, inRight);return root;}
}
Leetcode 297.二叉树的序列化与反序列化
Leetcode 297.二叉树的序列化与反序列化
使用先序遍历来构成数组,数组转换成树。
public class Codec {public String serialize(TreeNode root) {return rserialize(root, "");}private String rserialize(TreeNode root, String s) {if (root == null) {s += "NONE,";} else {s += root.val + ",";s = rserialize(root.left, s);s = rserialize(root.right, s);}return s;}// Decodes your encoded data to tree.public TreeNode deserialize(String data) {String[] dataList = data.split(",");return rdeserialize(new ArrayList(Arrays.asList(dataList)));}private TreeNode rdeserialize(List<String> dataList) {if (dataList.get(0).equals("NONE")) {dataList.remove(0);return null;}TreeNode root = new TreeNode(Integer.valueOf(dataList.get(0)));dataList.remove(0);root.left = rdeserialize(dataList);root.right = rdeserialize(dataList);return root;}
}
二叉树的递归思维
LeetCode572.另一个树的子树
LeetCode572.另一个树的子树
简化成两棵树是否相等
class Solution {public boolean isSubtree(TreeNode root, TreeNode subRoot) {if (root == null) {return false;}return isSameTree(root, subRoot) || isSubtree(root.left, subRoot) || isSubtree(root.right, subRoot);}public boolean isSameTree(TreeNode s, TreeNode t) {if (s == null && t == null) {return true;}if (s == null || t == null || s.val != t.val) {return false;}return isSameTree(s.left, t.left) && isSameTree(s.right, t.right);}
}
其他
LeetCode114. 二叉树展开为链表
比较简单的方法就是先序遍历元素。
class Solution {List<TreeNode> list = new ArrayList<>();public void flatten(TreeNode root) {dfs(root);for (int i = 1; i < list.size(); i++) {TreeNode prev = list.get(i - 1), cur = list.get(i);prev.left = null;prev.right = cur;}}private void dfs(TreeNode root) {if (root == null) {return;}list.add(root);dfs(root.left);dfs(root.right);}
}
把右节点放在左节点的最右节点的右边,把左节点放在右边,这样顺序是正确的。
class Solution {public void flatten(TreeNode root) {TreeNode cur = root;while (cur != null) {TreeNode left = cur.left;if (left != null) {TreeNode tmp = left;while (tmp.right != null) {tmp = tmp.right;}tmp.right = cur.right;cur.left = null;cur.right = left;}cur = cur.right;}}
}