110 Balanced Binary Tree

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博客围绕判断二叉树是否为高度平衡树展开。高度平衡二叉树定义为每个节点的左右子树深度差不超过1。分析了判断方法，即计算每个节点为根的树的高度，遍历节点看是否平衡，还提及了尝试解和标准解。

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1 题目

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 depth of the two subtrees of everynode never differ by more than 1.

Example 1:

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

Return true.

Example 2:

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

Return false.

2 尝试解

2.1 分析

判断是否为平衡二叉树，即任意节点左右子树高度相差不差过2。可以让每一个节点的为以节点为根节点的树的高度，然后遍历每个节点，看其是否平衡。

2.2 代码

class Solution {
public:
    bool isBalanced(TreeNode* root) {
        if(root == NULL) return true;
        depth(root);
        queue<TreeNode*> traverse;
        traverse.push(root);
        while(!traverse.empty()){
            TreeNode* cur = traverse.front();
            traverse.pop();
            if(abs((cur->left?cur->left->val:0)-(cur->right?cur->right->val:0))>=2) return false;
            if(cur->left != NULL) traverse.push(cur->left);
            if(cur->right != NULL) traverse.push(cur->right); 
        }
        return true;
    }
    int depth(TreeNode* root){
        if(root == NULL) return 0;
        if(root->left == NULL && root->right == NULL) {
            root->val = 1;
            return 1;
        }
        root->val = 1 + max(depth(root->left),depth(root->right));
        return root->val;
    }
};

3 标准解

class solution {
public:
    int depth (TreeNode *root) {
        if (root == NULL) return 0;
        return max (depth(root -> left), depth (root -> right)) + 1;
    }

    bool isBalanced (TreeNode *root) {
        if (root == NULL) return true;
        
        int left=depth(root->left);
        int right=depth(root->right);
        
        return abs(left - right) <= 1 && isBalanced(root->left) && isBalanced(root->right);
    }
};