一、LeetCode 110 平衡二叉树
题目链接: 110.平衡二叉树
https://leetcode.cn/problems/balanced-binary-tree/
思路：设置深度计算函数，进行递归处理。
class Solution {public boolean isBalanced(TreeNode root) {if(root == null){return true;}boolean left = isBalanced(root.left);boolean right = isBalanced(root.right);int length_left = depth(root.left);int length_right = depth(root.right);int flag = length_left - length_right;if(flag >= -1 && flag <= 1){//本层平衡且左右子树均平衡return true && left && right;}else{return false;}}//计算子树深度public int depth(TreeNode root){if(root == null){return 0;}int left = depth(root.left);int right = depth(root.right);return left>right ? left+1 : right+1;}
}
/*** Definition for a binary tree node.* public class TreeNode {* int val;* TreeNode left;* TreeNode right;* TreeNode() {}* TreeNode(int val) { this.val = val; }* TreeNode(int val, TreeNode left, TreeNode right) {* this.val = val;* this.left = left;* this.right = right;* }* }*/二、LeetCode 257 二叉树的所有路径
题目链接：257.二叉树的所有路径https://leetcode.cn/problems/binary-tree-paths/description/
思路：递归+回溯。
class Solution {public List<String> binaryTreePaths(TreeNode root) {List<Integer> path = new ArrayList<>();List<String> ans = new ArrayList<>();path(root,path,ans);return ans;}public void path(TreeNode root, List<Integer> path, List<String> ans){path.add(root.val); //中//若为叶子节点则输出该路径if(root.left == null && root.right == null){StringBuilder sb = new StringBuilder();for(int i = 0; i < path.size()-1; i++){sb.append(path.get(i)).append("->");}//加入叶子节点sb.append(root.val);ans.add(sb.toString());}//回溯处理if(root.left != null){path(root.left,path,ans);path.remove(path.size()-1);}if(root.right != null){path(root.right,path,ans);path.remove(path.size()-1);}}
}
/*** Definition for a binary tree node.* public class TreeNode {* int val;* TreeNode left;* TreeNode right;* TreeNode() {}* TreeNode(int val) { this.val = val; }* TreeNode(int val, TreeNode left, TreeNode right) {* this.val = val;* this.left = left;* this.right = right;* }* }*/三、LeetCode 404 左叶子之和
题目链接：404.左叶子之和https://leetcode.cn/problems/sum-of-left-leaves/
思路：后序递归遍历、左叶子节点处理。
class Solution {public int sumOfLeftLeaves(TreeNode root) {if(root == null){return 0;}//递归遍历 左右中int left = sumOfLeftLeaves(root.left);int right = sumOfLeftLeaves(root.right);int mid = 0;//遇到左叶子节点就记录数值，否则记为0if(root.left != null && root.left.left == null && root.left.right == null){mid = root.left.val;}return left + mid + right;}
}
/*** Definition for a binary tree node.* public class TreeNode {* int val;* TreeNode left;* TreeNode right;* TreeNode() {}* TreeNode(int val) { this.val = val; }* TreeNode(int val, TreeNode left, TreeNode right) {* this.val = val;* this.left = left;* this.right = right;* }* }*/四、小结
假期结束，开刷OVO
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