LeetCode | Balanced Binary Tree
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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 every node never differ by more than 1.
判断平衡二叉树,常规的写法是两个递归,一个递归用于计算所有节点,另一个递归用于计算每一个节点的深度。
新的方法是直接使用递归计算深度,但是在计算深度之前需要判断左右子树的深度之差是否大于1,如果是,说明不符合要求,以-1标记。递归下来的时候,只要计算到left、right或者当前的abs(left-right)>1的时候就表明子树或者自己本身不符合平衡二叉树,递归返回。
class Solution {public: bool isBalanced(TreeNode* root) { return isBalancedTree(root) >=0; } int isBalancedTree(TreeNode* root){ if(!root) return 0; int left=isBalancedTree(root->left); int right=isBalancedTree(root->right); //判断不符合,提前退出 //类似if(abs(left-right)>1) return false;这句 //也就是这个条件不符合就return false //符合这个条件了,还需要继续判断后面的 if(left<0 || right<0 || abs(left-right)>1) return -1; return left>right?(left+1):(right+1); } // bool isBalanced(TreeNode* root) { // if(root==NULL) return true; // //只有一个bool,深度如何可以传递 // //事实证明需要另开一个函数 // int left=getDepth(root->left); // int right=getDepth(root->right); // if(abs(left-right)>1) return false; // //于是这种还是利用了双循环 // return isBalanced(root->left) && isBalanced(root->right); // } // int getDepth(TreeNode* root){ // int depth=0; // if(root){ // int left=getDepth(root->left); // int right=getDepth(root->right); // return left>right?(left+1):(right+1); // } // return depth; // }};
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