leetcode617. 合并二叉树

1.题目描述：

给定两个二叉树，想象当你将它们中的一个覆盖到另一个上时，两个二叉树的一些节点便会重叠。你需要将他们合并为一个新的二叉树。合并的规则是如果两个节点重叠，那么将他们的值相加作为节点合并后的新值，否则不为NULL的节点将直接作为新二叉树的节点。

2.递归：

深度优先搜索，mergeTrees(TreeNode root1, TreeNode root2)方法中有两个形参，解题构建一颗新的二叉树，则递归return返回值为新二叉树的节点。递归三步骤，①递归结束条件：root1和root2均为null，返回null。②每级递归需要做的：判断合并的节点是否为空，为空则返回非空节点，被上一级递归调用后所合并（这个也可以算到第一步骤中）。合并的节点均为非空，则新建节点存储和，再通过递归添加新建节点的左右子节点。③返回值：每级递归都返回合并后的新节点。时间和空间复杂度均为O(min(m,n))，空间复杂度与递归调用的层数即树的高度在树为链表时最差。

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public TreeNode mergeTrees(TreeNode root1, TreeNode root2) {
        if(root1 == null && root2 == null) return null;//可以省略
        if(root1 == null) return root2;
        if(root2 == null) return root1;
        TreeNode newNode = new TreeNode(root1.val + root2.val);
        newNode.left = mergeTrees(root1.left,root2.left);
        newNode.right = mergeTrees(root1.right,root2.right);
        return newNode;
    }
}

当然也可以在原树上修改，节省空间并省略第一句递归结束条件：

class Solution {
    public TreeNode mergeTrees(TreeNode root1, TreeNode root2) {
        if(root1 == null) return root2;
        if(root2 == null) return root1;
        root1.val += root2.val;
        root1.left = mergeTrees(root1.left,root2.left);
        root1.right = mergeTrees(root1.right,root2.right);
        return root1;
    }
}

3.广度优先搜索迭代：此题较复杂，直接复制粘贴官解代码：

class Solution {
    public TreeNode mergeTrees(TreeNode t1, TreeNode t2) {
        if (t1 == null) return t2;
        if (t2 == null) return t1;
        TreeNode merged = new TreeNode(t1.val + t2.val);
        Queue<TreeNode> queue = new LinkedList<TreeNode>();
        Queue<TreeNode> queue1 = new LinkedList<TreeNode>();
        Queue<TreeNode> queue2 = new LinkedList<TreeNode>();
        queue.offer(merged);
        queue1.offer(t1);
        queue2.offer(t2);
        while (!queue1.isEmpty() && !queue2.isEmpty()) {
            TreeNode node = queue.poll(), node1 = queue1.poll(), node2 = queue2.poll();
            TreeNode left1 = node1.left, left2 = node2.left, right1 = node1.right, right2 = node2.right;
            if (left1 != null || left2 != null) {
                if (left1 != null && left2 != null) {
                    TreeNode left = new TreeNode(left1.val + left2.val);
                    node.left = left;
                    queue.offer(left);
                    queue1.offer(left1);
                    queue2.offer(left2);
                } else if (left1 != null) {
                    node.left = left1;
                } else if (left2 != null) {
                    node.left = left2;
                }
            }
            if (right1 != null || right2 != null) {
                if (right1 != null && right2 != null) {
                    TreeNode right = new TreeNode(right1.val + right2.val);
                    node.right = right;
                    queue.offer(right);
                    queue1.offer(right1);
                    queue2.offer(right2);
                } else if (right1 != null) {
                    node.right = right1;
                } else {
                    node.right = right2;
                }
            }
        }
        return merged;
    }
}