LeetCode——Merge Two Binary Trees

LeetCode——Merge Two Binary Trees

# 617

Given two binary trees and imagine that when you put one of them to cover the other, some nodes of the two trees are overlapped while the others are not.

You need to merge them into a new binary tree. The merge rule is that if two nodes overlap, then sum node values up as the new value of the merged node. Otherwise, the NOT null node will be used as the node of new tree.

Example 1:

Input:     Tree 1                     Tree 2                            1                         2                                      / \                       / \                                    3   2                     1   3                               /                           \   \                            5                             4   7                  Output: Merged tree:         3        / \       4   5      / \   \      5   4   7

Note: The merging process must start from the root nodes of both trees.

这一题的目的是合并两个二叉树，如果两个节点不空，则新节点的值为两者之和，否则非空节点作为新节点。这一题显然是可以用递归写的，也很简单。

• C++

/** * Definition for a binary tree node. * struct TreeNode { *     int val; *     TreeNode *left; *     TreeNode *right; *     TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */class Solution {public:    TreeNode* mergeTrees(TreeNode* t1, TreeNode* t2) {        if(!t1 && !t2)            return 0;        if(!t1 && t2)            return t2;        if(t1 && !t2)            return t1;        t1 -> val = t1 -> val + t2 -> val;        t1 -> left = mergeTrees(t1 -> left,t2 -> left);        t1 -> right = mergeTrees(t1 -> right,t2 -> right);        return t1;    }};

reveal solution里面也只是用了递归和迭代的方法，迭代的方法可以由递归的方法转换而来。这里使用了一个栈来存储节点，以此实现迭代。

• Java

/** * Definition for a binary tree node. * public class TreeNode { *     int val; *     TreeNode left; *     TreeNode right; *     TreeNode(int x) { val = x; } * } */public class Solution {    public TreeNode mergeTrees(TreeNode t1, TreeNode t2) {        if (t1 == null)            return t2;        Stack < TreeNode[] > stack = new Stack < > ();        stack.push(new TreeNode[] {t1, t2});        while (!stack.isEmpty()) {            TreeNode[] t = stack.pop();            if (t[0] == null || t[1] == null) {                continue;            }            t[0].val += t[1].val;            if (t[0].left == null) {                t[0].left = t[1].left;            } else {                stack.push(new TreeNode[] {t[0].left, t[1].left});            }            if (t[0].right == null) {                t[0].right = t[1].right;            } else {                stack.push(new TreeNode[] {t[0].right, t[1].right});            }        }        return t1;    }}