LeetCode 0110 Balanced Binary Tree【平衡二叉树，dfs】

Given a binary tree, determine if it is height-balanced.

For this problem, a height-balanced binary tree is defined as:

a binary tree in which the left and right subtrees of every node differ in height by no more than 1.

Example 1:

Given the following tree [3,9,20,null,null,15,7]:

    3
   / \
  9  20
    /  \
   15   7

Return true.

Example 2:

Given the following tree [1,2,2,3,3,null,null,4,4]:

       1
      / \
     2   2
    / \
   3   3
  / \
 4   4

Return false.

---

题意

判断二叉树是否是平衡二叉树

思路1

• 满足左右子树深度差不超过1，左子树是平衡的，右子树是平衡的三个条件

代码1

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    bool isBalanced(TreeNode* root) {
        if(root == NULL) return true;
        int depth_left = dfs(root->left);
        int depth_right = dfs(root->right);        
        return (abs(depth_left-depth_right) <= 1) && isBalanced(root->left) && isBalanced(root->right);
    }
    
    int dfs(TreeNode *node){
        if(node == NULL) return 0;
        int left = dfs(node->left);
        int right = dfs(node->right);
        return left > right ? (left + 1) : (right + 1);
    }
};

递归求二叉树深度

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int dfs(TreeNode *node){
    if(node == NULL) return 0;
    int left = dfs(node->left);
    int right = dfs(node->right);
    return left > right ? (left + 1) : (right + 1);
}

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
private:
    bool ans = true;
public:
    bool isBalanced(TreeNode* root) {
        maxDepth(root);
        return ans;
    }
    
    int maxDepth(TreeNode *node) {
        if(node == NULL) return 0;
        int l = maxDepth(node->left);
        int r = maxDepth(node->right);
        if(abs(l - r) > 1) ans = false;
        return 1 + max(l, r);
    }
};