LintCode | 93. 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。
题目链接

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递归解决，递归函数每次返回两棵子数的深度，并进行比较，相等或相差仅为1则返回最大值，否则返回-1。用-1表示该子树已不平衡。

/** * 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) {        if(getDepth(root) == -1) return false;        else return true;    }    private int getDepth(TreeNode node) {        if(node == null) return 0;        int rightDepth = getDepth(node.right);        int leftDepth = getDepth(node.left);        if(leftDepth == -1 || rightDepth == -1) return -1;        else if(rightDepth == leftDepth) return rightDepth + 1;        else if(rightDepth - leftDepth == 1) return rightDepth + 1;        else if(leftDepth - rightDepth == 1) return leftDepth + 1;        else return -1;    }}