LeetCode——Find Bottom Left Tree Value

Question

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.

Solution

先求出樹的高度，而後層次遍歷，遍歷到最後一層，輸出最左邊的節點(也就是最後一層第一個節點，由於層次遍歷採用從左到右的方式)。

Code

/**
 * 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) {
        int height = calcTreeHeight(root);
        int count = 1;
        queue<TreeNode*> childs, new_childs;
        childs.push(root);
        while (1) {
            while(!childs.empty()) {
                TreeNode* tmp = childs.front();
                if (count == height)
                    return tmp->val;
                childs.pop();
                if (tmp->left != NULL)
                    new_childs.push(tmp->left);
                if (tmp->right != NULL)
                    new_childs.push(tmp->right);
            }   
            count++;
            childs = new_childs;
            clear(new_childs);
        }
        
    }
    // 清空隊列的方法
    void clear( std::queue<TreeNode*> &q )
    {
       std::queue<TreeNode*> empty;
       std::swap(q, empty );
    }
    int calcTreeHeight(TreeNode* root) {
        if (root == NULL)
            return 0;
        int left = calcTreeHeight(root->left);
        int right = calcTreeHeight(root->right);
        return max(left, right) + 1;
    }
};

Solution 2

直接層次遍歷也能夠，最後一層遍歷結束後，直接輸出最後一層的第一個節點便可。

Code 2

/**
 * 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) {
        queue<TreeNode*> childs, new_childs;
        childs.push(root);
        while (1) {
            int flag = 1;
            TreeNode* first;
            while(!childs.empty()) {
                TreeNode* tmp = childs.front();
                if (flag) {
                    first = tmp;
                    flag = 0;
                } 
                childs.pop();
                if (tmp->left != NULL)
                    new_childs.push(tmp->left);
                if (tmp->right != NULL)
                    new_childs.push(tmp->right);
            }
            if (new_childs.empty()) {
                return first->val;
            }
            childs = new_childs;
            clear(new_childs);
        }
        
    }
    void clear( std::queue<TreeNode*> &q )
    {
       std::queue<TreeNode*> empty;
       std::swap( q, empty );
    }
};