【LeetCode】Balanced Binary Tree 解题报告

【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 every node never differ by more than 1.

【说明】

不是很难，思路大家可能都会想到用递归，分别判断左右两棵子树是不是平衡二叉树，如果都是并且左右两颗子树的高度相差不超过1，那么这棵树就是平衡二叉树。

【比较直观的Java代码】

/** * Definition for binary tree * public class TreeNode { *     int val; *     TreeNode left; *     TreeNode right; *     TreeNode(int x) { val = x; } * } */public class Solution {    public boolean isBalanced(TreeNode root) {        if(root == null){            return true;        }        if(root.left==null && root.right==null){            return true;        }        if(Math.abs(depth(root.left)-depth(root.right)) > 1){            return false;        }        return isBalanced(root.left) && isBalanced(root.right);    }        public int depth(TreeNode root){        if(root==null){            return 0;        }        return 1+Math.max(depth(root.left), depth(root.right));    }}

【改进后】

/** * Definition for binary tree * public class TreeNode { *     int val; *     TreeNode left; *     TreeNode right; *     TreeNode(int x) { val = x; } * } */public class Solution {    public boolean isBalanced(TreeNode root) {        height(root);        return run(root);    }        public boolean run(TreeNode root) {        if (root == null) return true;                int l = 0, r = 0;        if (root.left != null) l = root.left.val;        if (root.right != null) r = root.right.val;        if (Math.abs(l - r) <= 1 && isBalanced(root.left) && isBalanced(root.right)) return true;                return false;    }        public int height(TreeNode root) {        if (root == null) return 0;        root.val = Math.max( height(root.left), height(root.right) ) + 1;         return root.val;    }}

利用了TreeNode结构中的val，用它来记录以当前结点为根的子树的高度，避免多次计算。