链接：

https://leetcode.com/problems/merge-two-binary-trees/

大意：

合并两棵树，合并规则如下：

（1）若两棵树对应位置都为null，则新树在该位置为null

（2）若两棵树中只有一棵树在对应位置为null，则新树在该位置的值为那个在该位置非空的值

（3）若两棵树在对应位置都不为null，则新树在该位置的值为两棵树在该位置的值的和

例子：

思路：

咋看，思路很多：递归构造树，层次遍历构造树，在原树（较高的一棵树）的基础上添加节点构造新树....

最终选择了递归构造树（毕竟只要是树，优先方法就是递归...）。递归方法的参数为：第一棵树的当前节点，第二棵树的当前节点，新树的父节点，标志域（标明是父节点的左孩子还是右孩子）。

代码：

class Solution {
    public TreeNode mergeTrees(TreeNode t1, TreeNode t2) {
        if(t1 == null)
            return t2;
        if(t2 == null)
            return t1;
        TreeNode res = new TreeNode(t1.val + t2.val), r = res;
        preOrderConstructTree(t1.left, t2.left, r, 0);
        preOrderConstructTree(t1.right, t2.right, r, 1);
        return res;
    }
    // tag=0:构建左子树  tag=1:构建右子树
    public void preOrderConstructTree(TreeNode r1, TreeNode r2, TreeNode f, int tag) {
        if (r1 == null && r2 == null) {
            return ;
        }
        if (r1 == null) {
            if (tag == 0) {
                f.left = new TreeNode(r2.val);
                preOrderConstructTree(null, r2.left, f.left, 0);
                preOrderConstructTree(null, r2.right, f.left, 1);
            }
            else {
                f.right = new TreeNode(r2.val);
                preOrderConstructTree(null, r2.left, f.right, 0);
                preOrderConstructTree(null, r2.right, f.right, 1);    
            }
        } else if (r2 == null) {
            if (tag == 0) {
                f.left = new TreeNode(r1.val);
                preOrderConstructTree(r1.left, null, f.left, 0);
                preOrderConstructTree(r1.right, null, f.left, 1);
            }
            else {
                f.right = new TreeNode(r1.val);
                preOrderConstructTree(r1.left, null, f.right, 0);
                preOrderConstructTree(r1.right, null, f.right, 1);   
            }
        } else if (tag == 0) {
            f.left = new TreeNode(r1.val + r2.val);
            preOrderConstructTree(r1.left, r2.left, f.left, 0);
            preOrderConstructTree(r1.right, r2.right, f.left, 1);
        } else {
            f.right = new TreeNode(r1.val + r2.val);
            preOrderConstructTree(r1.left, r2.left, f.right, 0);
            preOrderConstructTree(r1.right, r2.right, f.right, 1);
        }
    }
}

结果：

结论：

从结果看来，代码还是有需要改进的地方。初步判定是在当只有一个节点为null的情况下，我仍旧去遍历非null的树，并创建新的树节点。也许可以不遍历非null的树，而是直接把对应父节点的左（右）孩子指向非null树的当前节点。

最佳：

class Solution {
    public TreeNode mergeTrees(TreeNode t1, TreeNode t2) {
        if(t1 == null)
            return t2;
        if(t2 == null)
            return t1;
        TreeNode res = new TreeNode(t1.val + t2.val), r = res;
        preOrderConstructTree(t1.left, t2.left, r, 0);
        preOrderConstructTree(t1.right, t2.right, r, 1);
        return res;
    }
    // tag=0:构建左子树  tag=1:构建右子树
    public void preOrderConstructTree(TreeNode r1, TreeNode r2, TreeNode f, int tag) {
        if (r1 == null && r2 == null) {
            return ;
        }
        if (r1 == null) {
            if (tag == 0) {
                // f.left = new TreeNode(r2.val);
                // preOrderConstructTree(null, r2.left, f.left, 0);
                // preOrderConstructTree(null, r2.right, f.left, 1);
                f.left = r2;
            }
            else {
                // f.right = new TreeNode(r2.val);
                // preOrderConstructTree(null, r2.left, f.right, 0);
                // preOrderConstructTree(null, r2.right, f.right, 1);
                f.right = r2;
            }
        } else if (r2 == null) {
            if (tag == 0) {
                // f.left = new TreeNode(r1.val);
                // preOrderConstructTree(r1.left, null, f.left, 0);
                // preOrderConstructTree(r1.right, null, f.left, 1);
                f.left = r1;
            }
            else {
                // f.right = new TreeNode(r1.val);
                // preOrderConstructTree(r1.left, null, f.right, 0);
                // preOrderConstructTree(r1.right, null, f.right, 1);   
                f.right = r1;
            }
        } else if (tag == 0) {
            f.left = new TreeNode(r1.val + r2.val);
            preOrderConstructTree(r1.left, r2.left, f.left, 0);
            preOrderConstructTree(r1.right, r2.right, f.left, 1);
        } else {
            f.right = new TreeNode(r1.val + r2.val);
            preOrderConstructTree(r1.left, r2.left, f.right, 0);
            preOrderConstructTree(r1.right, r2.right, f.right, 1);
        }
    }
}