Same Tree 题解

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题目来源：

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Description

Given two binary trees, write a function to check if they are the same or not.

Two binary trees are considered the same if they are structurally identical and the nodes have the same value.

Example

Example 1:

Input:     1         1          / \       / \         2   3     2   3        [1,2,3],   [1,2,3]Output: true

Example 2:

Input:     1         1          /           \         2             2        [1,2],     [1,null,2]Output: false

Example 3:

Input:     1         1          / \       / \         2   1     1   2        [1,2,1],   [1,1,2]Output: false

Solution

class Solution {private:    bool isSameNode(TreeNode* node1, TreeNode* node2) {        if (node1 == NULL && node2 == NULL)            return true;        if (node1 != NULL && node2 != NULL) {            return (node1 -> val == node2 -> val) &&                    isSameNode(node1 -> left, node2 -> left) &&                    isSameNode(node1 -> right, node2 -> right);        } else {            return false;        }    }public:    bool isSameTree(TreeNode* node1, TreeNode* node2) {        return isSameNode(node1, node2);    }};

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解题描述

这道题是要判断给出的两棵二叉树是否完全相等，也就是判断两棵树是不是结构完全相同而且每个节点的数字是否完全一样。一开始我想到的还是简单的递归做法，也就是上面给出来的。不过跑出来的时间确实还是比较长。

AC之后看了讨论区，发现还是可以用非递归的方式来解决的。主要是使用栈来模拟递归调用，跑出来的时间也确实缩短了一些：

class Solution {public:    bool isSameTree(TreeNode* node1, TreeNode* node2) {        if (node1 == NULL && node2 == NULL)            return true;        if (node1 != NULL && node2 != NULL) {            stack
    
      stack1;            stack
     
       stack2;            stack1.push(node1);            stack2.push(node2);            TreeNode *n1, *n2;            while (!stack1.empty() && !stack2.empty()) {                n1 = stack1.top();                stack1.pop();                n2 = stack2.top();                stack2.pop();                if (n1 -> val != n2 -> val) return false;                if (n1 -> left != NULL)                    stack1.push(n1 -> left);                if (n2 -> left != NULL)                    stack2.push(n2 -> left);                if (stack1.size() != stack2.size())                    return false;                if (n1 -> right != NULL)                    stack1.push(n1 -> right);                if (n2 -> right != NULL)                    stack2.push(n2 -> right);                if (stack1.size() != stack2.size())                    return false;            }            return stack1.size() == stack2.size();        } else {            return false;        }    }};