Balanced Binary Tree

[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 ofevery node never differ by more than 1.

[题意]

   判断树是不是平衡二叉树

[分析]

  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

[实现]

/** * 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));    }}

<span style="font-size:18px;"></span><pre name="code" class="java">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。

/** * 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; } }