【leetcode】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.

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题解：

一种方法是写一个递归求高度的函数，然后再写一个递归函数判断树是否是平衡的。

代码如下：

public class Solution {
    private int height(TreeNode root){
        if(root == null)
            return 0;
        int left = height(root.left);
        int right = height(root.right);
        
        return Math.max(left, right)+1;
    }
    public boolean isBalanced(TreeNode root) {
        if(root == null)
            return true;
        int left = height(root.left);
        int right = height(root.right);
        
        if(Math.abs(left-right)>1)
            return false;
        return isBalanced(root.left) & isBalanced(root.right);
    }
}

这种方法耗时428ms。

第二种方法在递归求树的高度的过程中顺便判断树是否平衡，如果在某个节点处，该节点的左子树和右子树高度只差大于1，或者该树的左子树或者又子树不平衡，那么返回该树的高度为-1；否则返回该树的高度。

代码如下：

public class Solution {
    private int height(TreeNode root){
        if(root == null)
            return 0;
        int left = height(root.left);
        int right = height(root.right);
        
        if(left == -1 || right == -1 || Math.abs(left - right) > 1)
            return -1;
        return Math.max(left, right)+1;
    }
    public boolean isBalanced(TreeNode root) {
        return height(root) != -1;
    }
}

这种方法耗时464ms。

第二遍刷leetcode时候的java代码也放上来，耗时250ms。

public class Solution {
    public boolean isBalanced(TreeNode root) {
        height(root);
        return isBlanced;
    }
    boolean isBlanced = true;
    private int height(TreeNode root){
        if(root == null)
            return 0;
        int leftheight = height(root.left);
        int rightheight = height(root.right);
        if(Math.abs(leftheight - rightheight) > 1){
            isBlanced = false;
        }
        return Math.max(leftheight, rightheight)+1;
    }
}