6.leetCode617:Merge Two Binary Trees

题目：
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.

思路1：可以新建一棵树M，为A和B的合并树。M的每个节点值是A、B对应节点的值之和。若某一棵树对应节点为空，则M种对应节点的值为另一棵树的节点值。

/** * 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)     {        TreeNode root;        TreeNode left_1 = null, left_2 = null;        TreeNode right_1 = null, right_2 = null;        if(t1 == null && t2 == null)            return null;        else if(t1 != null && t2 != null)        {            root = new TreeNode(t1.val + t2.val);            left_1 = t1.left;            left_2 = t2.left;            right_1 = t1.right;            right_2 = t2.right;        }        else if(t1 != null && t2 == null)        {            root = new TreeNode(t1.val);            left_1 = t1.left;            right_1 = t1.right;        }        else        {            root = new TreeNode(t2.val);            left_2 = t2.left;            right_2 = t2.right;        }        root.left = mergeTrees(left_1, left_2);        root.right = mergeTrees(right_1, right_2);        return root;    }}

思路2：要合并两棵树A和B，以A为基础，将A、B中对应的节点的值进行相加，并更新A的节点值。

class Solution {    public TreeNode mergeTrees(TreeNode t1, TreeNode t2) {        if(t1 ==null && t2 ==null)            return null;        else if(t1 == null)            return t2;        else if(t2 == null)            return t1;        t1.val +=t2.val;        t1.left = mergeTrees(t1.left,t2.left);        t1.right = mergeTrees(t1.right,t2.right);        return t1;    }}

思路3：使用一个栈来存储对应的节点对。从根节点开始，若对应节点都不为空，则相加；若一对对应节点中的其中一个节点为空，则为不为空的节点值。当前节点处理完后，把其左孩子节点和右孩子节点的对应节点对都放入栈中。

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;    }}