剑指offer（36）-平衡二叉树

题目描述

---

输入一棵二叉树，判断该二叉树是否是平衡二叉树。

代码

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需要遍历结点多次的解法。

class Solution {
public:
    bool IsBalanced_Solution(TreeNode* pRoot) {
        if (pRoot == NULL) {
            return true;
        }
        int leftDepth = TreeDepth(pRoot->left);
        int rightDepth = TreeDepth(pRoot->right);

        int diff = leftDepth - rightDepth;
        if (diff > 1 || diff < -1) {
            return false;
        }
        return IsBalanced_Solution(pRoot->left) && IsBalanced_Solution(pRoot->right);
    }

    int TreeDepth(TreeNode *pRoot){
        if (pRoot == NULL) {
            return 0;
        }
        int leftDepth = TreeDepth(pRoot->left);
        int rightDepth = TreeDepth(pRoot->right);
        return (leftDepth > rightDepth) ? (leftDepth + 1): (rightDepth + 1);
    }

};

每个结点只遍历一次的解法

---

class Solution {
public:
    bool IsBalanced_Solution(TreeNode* pRoot) {
        int depth = 0;
        return Is_Balanced(pRoot, &depth);
    }

     bool Is_Balanced(TreeNode *pRoot, int *pDepth){
        if (pRoot == NULL) {
            *pDepth = 0;
            return true;
        }

        int left,right;
        if (Is_Balanced(pRoot->left, &left) && Is_Balanced(pRoot->right, &right)) {
            int diff = left - right;
            if (diff <= 1 && diff >= -1) {
                *pDepth = 1 + ((left > right) ? left : right);
                return true;
            }
        }
        return false;
    }


};