[Leetcode] 513. Find Bottom Left Tree Value 解题报告

题目：

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.

思路：

遇到树的问题，一般都要想着用DFS或者BFS，除非题目明确说明不允许。在本题目中，我们首先计算左子树的高度以及右子树的高度，如果右子树的高度大于左子树的高度，那么就返回右子树最左下的结点的值；否则就返回左子树最左下的结点的值。当然需要注意结点如果是叶子结点的特殊情况，这是递归的基础。

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) {
        int height = 0;
        return findBottomLeftValue(root, height);
    }
private:
    int findBottomLeftValue(TreeNode* root, int &height) {
        if (!root->left && !root->right) {      // a leaf
            height = 1;
            return root->val;
        }
        else {
            int max_height = 0, value = 0;
            if (root->left) {
                value = findBottomLeftValue(root->left, max_height);
            }
            if (root->right) {
                int right_height = 0;
                int right_value = findBottomLeftValue(root->right, right_height);
                if (right_height > max_height) {
                    max_height = right_height;
                    value = right_value;
                }
            }
            height = max_height + 1;
            return value;
        }
    }
};