(二叉树 BFS) leetcode993. Cousins in Binary Tree

In a binary tree, the root node is at depth 0, and children of each depth knode are at depth k+1.

Two nodes of a binary tree are cousins if they have the same depth, but have different parents.

We are given the root of a binary tree with unique values, and the values x and y of two different nodes in the tree.

Return true if and only if the nodes corresponding to the values x and yare cousins.

Example 1:

Input: root = [1,2,3,4], x = 4, y = 3
Output: false

Example 2:

Input: root = [1,2,3,null,4,null,5], x = 5, y = 4
Output: true

Example 3:

Input: root = [1,2,3,null,4], x = 2, y = 3
Output: false

Note:

1. The number of nodes in the tree will be between 2 and 100.
2. Each node has a unique integer value from 1 to 100.

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额，这个题，怎么说的，弄懂了方法，就会发现这个题是比较简单的，是easy题是有道理的。要好好扎实BFS的基础。

这个题就是先把每个结点的父节点和它的位置求出，然后再比较，可以用hash。

C++代码：

/**
 * 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:
    bool isCousins(TreeNode* root, int x, int y) {
        queue<pair<TreeNode*,TreeNode*> >  q;
        q.push(make_pair(root,nullptr));
        int depth = 0;
        unordered_map<int,pair<TreeNode*,int> > mp; //TreeNode*指的是这个结点的父结点。第二个int指的是深度。
        while(!q.empty()){
            for(int i = q.size(); i > 0; i--){
                auto t = q.front();
                q.pop();
                mp[t.first->val] = make_pair(t.second,depth);
                if(t.first->left) q.push(make_pair(t.first->left,t.first));
                if(t.first->right) q.push(make_pair(t.first->right,t.first));
            }
            depth++;
        }
        auto tx = mp[x],ty = mp[y];
        return tx.first != ty.first && tx.second == ty.second;
    }
};

标签：Binary,TreeNode,int,Cousins,tree,depth,二叉树,root,first
来源： https://www.cnblogs.com/Weixu-Liu/p/10787493.html