package top100.top二叉树;import top100.TreeNode;import java.util.*;public class TOP {//94. 二叉树的中序遍历List<Integer> res = new ArrayList<>();public List<Integer> inorderTraversal(TreeNode root) {inorder(root);return res;}private void inorder(TreeNode root) {if (root != null) {inorder(root.left);res.add(root.val);inorder(root.right);}}//104. 二叉树的最大深度int maxDepth = 0;public int maxDepth(TreeNode root) {if (root == null) {return maxDepth;}dfsMaxDepth(root, 0);return maxDepth;}private void dfsMaxDepth(TreeNode root, int res) {if (root == null) {maxDepth = Math.max(maxDepth, res);} else {dfsMaxDepth(root.left, res + 1);dfsMaxDepth(root.right, res + 1);}}//226. 翻转二叉树public TreeNode invertTree(TreeNode root) {if (root == null) {return null;}TreeNode right = root.right;root.right = root.left;root.left = right;invertTree(root.left);invertTree(root.right);return root;}//101. 对称二叉树public boolean isSymmetric(TreeNode root) {if (root == null) {return true;}return isSymmetric(root.left, root.right);}public boolean isSymmetric(TreeNode left, TreeNode right) {if (left == null && right == null) {return true;} else if (left == null || right == null) {return false;} else if (left.val != right.val) {return false;} else {return isSymmetric(left.left, right.right) && isSymmetric(left.right, right.left);}}//543. 二叉树的直径int diameterOfBinaryTree = 0;public int diameterOfBinaryTree(TreeNode root) {dfsDiameterOfBinaryTree(root);return diameterOfBinaryTree;}//计算当前节点到其叶子节点的最长距离private int dfsDiameterOfBinaryTree(TreeNode root) {if (root == null) return 0;//左右节点的最长距离int l = dfsDiameterOfBinaryTree(root.left);int r = dfsDiameterOfBinaryTree(root.right);diameterOfBinaryTree = Math.max(diameterOfBinaryTree, l + r);return Math.max(l, r) + 1;}//102. 二叉树的层序遍历public List<List<Integer>> levelOrder(TreeNode root) {List<List<Integer>> res = new ArrayList<>();if (root == null) {return res;}Queue<TreeNode> queue = new LinkedList<>();queue.add(root);while (!queue.isEmpty()) {int n = queue.size();List<Integer> list = new ArrayList<>();for (int i = 0; i < n; i++) {TreeNode node = queue.poll();list.add(node.val);if (node.left != null) {queue.add(node.left);}if (node.right != null) {queue.add(node.right);}}res.add(list);}return res;}//108. 将有序数组转换为二叉搜索树public TreeNode sortedArrayToBST(int[] nums) {if (nums.length == 0) return null;int start = 0, mid = (nums.length - 1) / 2;TreeNode root = new TreeNode(nums[mid]); //二叉搜索树的根节点是中序遍历数组中间的数if (mid - start > 0) { //如果数组可以构成左子树root.left = sortedArrayToBST(Arrays.copyOfRange(nums, start, mid));}if (mid < nums.length - 1) { //如果数组可以构成右子树root.right = sortedArrayToBST(Arrays.copyOfRange(nums, mid + 1, nums.length));}return root;}//98. 验证二叉搜索树long preVal = Long.MIN_VALUE; //为了通过测试用例，设置比int最小还要小的值public boolean isValidBST(TreeNode root) {if (root == null) return true;//先判断左子树是否满足if (!isValidBST(root.left)) {return false;}//中序是否满足递增if (root.val <= preVal) {return false;}preVal = root.val; //更新pre值//判断又子树return isValidBST(root.right);}//230. 二叉搜索树中第K小的元素int res1;int k;ArrayList<Integer> list = new ArrayList<>();public int kthSmallest(TreeNode root, int k) {this.k = k;inOrder(root, k);return res1;}//中序遍历,从最小结点开始遍历:二叉搜索树的中序遍历为递增序列private void inOrder(TreeNode root, int k) {if (root == null) return;if (root.left != null) {inOrder(root.left, k);}list.add(root.val);if (list.size() == k) {res1 = root.val;return;}if (root.right != null) {inOrder(root.right, k);}}//230. 二叉搜索树中第K小的元素int res1;int k;public int kthSmallest(TreeNode root, int k) {this.k = k;inOrder(root);return res1;}private void inOrder(TreeNode root) {if (root == null) return;//先递归到左端点inOrder(root.left);if (k == 0) return;//遍历到第k个if (--k == 0) res1 = root.val;//中序遍历inOrder(root.right);}//199. 二叉树的右视图public List<Integer> rightSideView(TreeNode root) {List<Integer> res = new ArrayList<>();if (root == null) {return res;}Queue<TreeNode> queue = new LinkedList<>();queue.add(root);while (!queue.isEmpty()) {int n = queue.size();for (int i = 0; i < n; i++) {TreeNode cur = queue.poll();if (i == n - 1) {res.add(cur.val);}if (cur.left != null) {queue.add(cur.left);}if (cur.right != null) {queue.add(cur.right);}}}return res;}//114. 二叉树展开为链表//方法1：逆前序遍历TreeNode preTreeNode = null;public void flatten(TreeNode root) {if (root == null) {return;}flatten(root.right);flatten(root.left);root.left = null;root.right = preTreeNode;preTreeNode = root;}//方法2：前序遍历后重组public void flatten(TreeNode root) {ArrayList<TreeNode> list = new ArrayList<>();preOder(root, list);for (int i = 1; i < list.size(); i++) {TreeNode pre = list.get(i - 1);TreeNode cur = list.get(i);pre.left = null;pre.right = cur;}}private void preOder(TreeNode root, ArrayList<TreeNode> list) {if (root != null) {list.add(root);if (root.left != null) {preOder(root.left, list);}if (root.right != null) {preOder(root.right, list);}}}//105. 从前序与中序遍历序列构造二叉树public TreeNode buildTree(int[] preorder, int[] inorder) {int n = preorder.length;if (n == 0) return null;TreeNode root = new TreeNode(preorder[0]);//获取根节点在中序的位置int mid = 0;for (int i = 0; i < n; i++) {if (inorder[i] == preorder[0]) {mid = i;break;}}//构造左右子树root.left = buildTree(Arrays.copyOfRange(preorder, 1, mid + 1), Arrays.copyOfRange(inorder, 0, mid));root.right = buildTree(Arrays.copyOfRange(preorder, mid + 1, preorder.length), Arrays.copyOfRange(inorder, mid + 1, inorder.length));return root;}//437. 路径总和 IIIint res4 = 0;//从每个节点dfspublic int pathSum(TreeNode root, int targetSum) {if (root == null) return 0;//遍历每个节点 bfsQueue<TreeNode> queue = new LinkedList<>();queue.add(root);while (!queue.isEmpty()) {int n = queue.size();for (int i = 0; i < n; i++) {TreeNode cur = queue.poll();dfs4(cur, targetSum);if (cur.left != null) queue.add(cur.left);if (cur.right != null) queue.add(cur.right);}}return res4;}private void dfs4(TreeNode root, long curSum) {if (root != null) {curSum -= root.val;if (curSum == 0) {res4++;}//递归dfs4(root.left, curSum);dfs4(root.right, curSum);}}//236. 二叉树的最近公共祖先public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {if(root == null || root == p || root == q) return root;TreeNode left = lowestCommonAncestor(root.left, p, q);TreeNode right = lowestCommonAncestor(root.right, p, q);if(left == null) return right;if(right == null) return left;return root;}//124. 二叉树中的最大路径和int res = Integer.MIN_VALUE;public int maxPathSum(TreeNode root) {dfs6(root);return res;}//计算root作为父结点所在在路径的最大路径和private int dfs6(TreeNode root) {if (root == null) return 0;//计算左子树的最大路径和,抛弃负数的路径int left = Math.max(0, dfs6(root.left));int right = Math.max(0, dfs6(root.right));//计算路径 left -> root -> right 的和res = Math.max(res, left + right + root.val);//返回root结点所在的子树最大路径和给父结点使用return root.val + Math.max(left, right);}
}