【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】

不是非常难，思路大家可能都会想到用递归，分别推断左右两棵子树是不是平衡二叉树，假设都是而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。

【比較直观的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。