(二叉树 BFS) leetcode513. Find Bottom Left Tree Value
Given a binary tree, find the leftmost value in the last row of the tree.
Example 1:
Input: 2 / \ 1 3 Output: 1
Example 2:
Input: 1 / \ 2 3 / / \ 4 5 6 / 7 Output: 7
Note: You may assume the tree (i.e., the given root node) is not NULL.
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这个题就是求出二叉树的最后一层的最左边的结点的数。
可以用BFS,也可以用DFS,不过BFS相对会简单一些的。
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: int findBottomLeftValue(TreeNode* root) { if(!root) return 0; queue<TreeNode*> q; q.push(root); int res = root->val; while(!q.empty()){ int len = q.size(); //必须加上这个,如果用i==q.size(),则会出错,因为q.size()会变化。 for(int i = len; i > 0; i--){ auto t = q.front(); q.pop(); if(i == len){ res = t->val; } if(t->left) q.push(t->left); if(t->right) q.push(t->right); } } return res; } };