题目

 平衡二叉树

给定一个二叉树,确定它是高度平衡的。对于这个问题,一棵高度平衡的二叉树的定义是:一棵二叉树中每个节点的两个子树的深度相差不会超过1。 

样例

给出二叉树 A={3,9,20,#,#,15,7}, B={3,#,20,15,7}

A)  3            B)    3 
   / \                  \
  9  20                 20
    /  \                / \
   15   7              15  7

二叉树A是高度平衡的二叉树,但是B不是

解题

递归求高度

判断左右孩子高度之差是小于等于1,这里还是需要递归的

/**
 * Definition of TreeNode:
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left, right;
 *     public TreeNode(int val) {
 *         this.val = val;
 *         this.left = this.right = null;
 *     }
 * }
 */
public class Solution {
    /**
     * @param root: The root of binary tree.
     * @return: True if this Binary tree is Balanced, or false.
     */
    public boolean isBalanced(TreeNode root) {
        // write your code here
        if(root == null)
            return true;
        int lft = depth(root.left);
        int rit = depth(root.right);
        if(Math.abs(lft-rit)<=1){
            return isBalanced(root.left)&&isBalanced(root.right);
        }
        else
            return false;
        
    }
    public int depth(TreeNode root){
        if(root ==null)
            return 0;
        if(root.left==null && root.right ==null)
            return 1;
        int lft = depth(root.left);
        int rit = depth(root.right);
        return Math.max(lft,rit) + 1;
    }
}
Java Code