[leedcode 110] Balanced Binary Tree

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.

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public boolean isBalanced(TreeNode root) {
        //需要构造一个函数求一个节点的高度，然后isBalanced函数每一层开始验证左右字子树是否满足要求，此种思想非常直观
        //这种解法效率不高，因为求每个节点是否满足时，需要求一遍高度再验证，然后再求下一层，中间计算有重复计算
        //另一种优化算法，在求高度时，就能把信息直接返回，不满足的默认直接回-1，不需要再比较
        if(root==null) return true;
        int left=getDepth(root.left);
        int right=getDepth(root.right);
        if(left>right+1||right>left+1) return false;
        else{
            return isBalanced(root.left)&&isBalanced(root.right);
        }
    }
    int getDepth(TreeNode root){
        if(root==null) return 0;
        return Math.max(getDepth(root.left),getDepth(root.right))+1;
        
    }
    
    /* public boolean isBalanced(TreeNode root){
       
       int temp=getDepth(root);
       return temp>-1;
    }
    public int getDepth(TreeNode root){
        if(root==null) return 0;
        int left=getDepth(root.left);
        int right=getDepth(root.right);
        int temp=Math.abs(left-right);
        if(left==-1||right==-1||temp>1) return -1;
        
       return  left>right?left+1:right+1;
    }*/
}