Leetcode 3701 · Find Nearest Right Node in Binary Tree (遍历和BFS好题)

3701 · Find Nearest Right Node in Binary TreePRE
Algorithms

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Description
Given a binary tree with a root node root and a node u in the tree, return the node value val of the right-hand node nearest to it in the layer in which the node u is located, or -1 if the node u is the rightmost node in the current layer.

The total number of nodes is between

All nodes in the tree have node values that are unique
u is a node in a binary tree rooted at root

Example
Example 1:

Input:
root = {1,2,3,#,4,5,6}
u = 2
Output:
3
Explanation:
As shown in the figure, in the layer where node 2 is located, the nearest right node is node 3
3701_1.png

Example 2:

Input:
root = {3,1,#,#,2}
u = 1
Output:
-1
Explanation:
As shown in the figure, the layer where node 1 is located has only one node and the nearest right node is not found

解法1：遍历
采用前序遍历。当找到目标节点后，记住当前层次。那么，前序遍历再次到达该层次的时候，访问到的节点就是所求节点。

/*** Definition of TreeNode:* class TreeNode {* public:*     int val;*     TreeNode *left, *right;*     TreeNode(int val) {*         this->val = val;*         this->left = this->right = NULL;*     }* }*/class Solution {
public:/*** @param root: The root of the binary tree* @param u: A node in root* @return: Node value of the right node*/int findNearestRightNode(TreeNode *root, TreeNode *u) {helper(root, 0, u);if (findNode) return findNode->val;return -1;}
private:bool find = false;int targetDepth = -1;TreeNode *findNode = NULL;void helper(TreeNode *root, int depth, TreeNode *u) {if (!root) return;if (findNode) return;if (find && depth == targetDepth) {findNode = root;return;}if (root == u) {find = true;targetDepth = depth;}helper(root->left, depth + 1, u);helper(root->right, depth + 1, u);}
};

写成这样也可以。这样就不用find变量了。

/*** Definition of TreeNode:* class TreeNode {* public:*     int val;*     TreeNode *left, *right;*     TreeNode(int val) {*         this->val = val;*         this->left = this->right = NULL;*     }* }*/class Solution {
public:/*** @param root: The root of the binary tree* @param u: A node in root* @return: Node value of the right node*/int findNearestRightNode(TreeNode *root, TreeNode *u) {helper(root, 0, u);if (findNode) return findNode->val;return -1;}
private:int targetDepth = -1;TreeNode *findNode = NULL;void helper(TreeNode *root, int depth, TreeNode *u) {if (!root || findNode) return;if (root == u) {targetDepth = depth;} else if (targetDepth == depth) {findNode = root;return;}helper(root->left, depth + 1, u);helper(root->right, depth + 1, u);}
};

注意: 下面这个写法不对。
只用find是不够的，因为还有些其它同层次的节点在处理，结果会覆盖resValue。
比如说输入：root = [1,2,3,null,4,5,6], u = 4
下面会输出：6。但结果应该是5。

class Solution {
public:/*** @param root: The root of the binary tree* @param u: A node in root* @return: Node value of the right node*/int findNearestRightNode(TreeNode *root, TreeNode *u) {helper(root, 0, u);return resValue;}
private:bool find = false;int resValue = -1, targetDepth = -1;void helper(TreeNode *root, int depth, TreeNode *u) {if (!root) return;if (find && depth == targetDepth) {resValue = root->val;return;}if (root == u) {find = true;targetDepth = depth;}helper(root->left, depth + 1, u);helper(root->right, depth + 1, u);}
};

解法2: bfs

/*** Definition of TreeNode:* class TreeNode {* public:*     int val;*     TreeNode *left, *right;*     TreeNode(int val) {*         this->val = val;*         this->left = this->right = NULL;*     }* }*/class Solution {
public:/*** @param root: The root of the binary tree* @param u: A node in root* @return: Node value of the right node*/int findNearestRightNode(TreeNode *root, TreeNode *u) {if (!root ||! u) return -1;queue<TreeNode *> q;int res = -1;bool find = false;q.push(root);while(!q.empty()) {int qSize = q.size();find = false;for (int i = 0; i < qSize; i++) {TreeNode *frontNode = q.front();q.pop();if (find) return frontNode->val;if (frontNode == u) find = true;if (frontNode->left) q.push(frontNode->left);if (frontNode->right) q.push(frontNode->right);}}return -1;}
};