LeetCode 513. Find Bottom Left Tree Value (C++)

Given a binary tree, find the leftmost value in the last row of the tree.

Example 1:

Input:    2   / \  1   3Output:1

Example 2: 

Input:        1       / \      2   3     /   / \    4   5   6       /      7Output:7

Note: You may assume the tree (i.e., the given root node) is not NULL.

思路：运用二叉树的层序遍历（宽度优先搜索），当到达最底层时，返回最底层中最左边节点的值。

一般来说，二叉树的层序遍历可以用queue来实现，从root节点开始，将非空的左右子节点压入queue中。但是这样难以确定是否已经到达最底层，需要额外增加一个queue记录对应节点所在的层数。

二叉树层序遍历的另外一种思路是用vector+两个标识来实现。类似queue的压栈做法，用两个标识cur和end来指示当前层的起始节点和结束节点的后一位。这样就可以通过这两个表示来判断是否已经到达最底层。代码如下

/** * 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) {        std::vector<TreeNode*> nVector;        int cur = 0;        int end = 0;        bool reach_bottom = false;        if (!root) return -1;        nVector.push_back(root);        while(cur < nVector.size()) {            end = nVector.size();            reach_bottom = true;            for(int i = cur;i < end; ++i) {                if(nVector[i]->left || nVector[i]->right) {                    reach_bottom = false;                    break;                }            }            if (reach_bottom) {                return nVector[cur]->val;            }            while (cur < end) {                if(nVector[cur]->left) nVector.push_back(nVector[cur]->left);                if(nVector[cur]->right) nVector.push_back(nVector[cur]->right);                ++cur;            }                    }        return -1;    }};

优化：其实判断是否到达底部是没有必要的，只需要增加一个cur_left节点用来存储每层最左边节点。当宽度优先搜索结束后，cur_left就是最底层最左边节点。代码如下：

class Solution {public:    int findBottomLeftValue(TreeNode* root) {        std::vector<TreeNode*> nVector;        int cur = 0;        int end = 0;        TreeNode* cur_left = NULL;        bool reach_bottom = false;        if (!root) return -1;        nVector.push_back(root);        while(cur < nVector.size()) {            end = nVector.size();            cur_left = nVector[cur];            while (cur < end) {                if(nVector[cur]->left) nVector.push_back(nVector[cur]->left);                if(nVector[cur]->right) nVector.push_back(nVector[cur]->right);                ++cur;            }                    }        return cur_left->val;    }};

另一种思路是进行深度优先搜索，记录深度，一旦发现现在的深度大于所记录的max depth，更新max depth和max depth value。（因为能够保证这时的节点是所在层最左边的节点）

/** * 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:    void findBottomLeftValue(TreeNode* root, int& maxDepth, int& leftVal, int depth) {        if (root == NULL) {            return;        }        //Go to the left and right of each node         findBottomLeftValue(root->left, maxDepth, leftVal, depth+1);        //Update leftVal and maxDepth        findBottomLeftValue(root->right, maxDepth, leftVal, depth+1);                //Update leftVal and maxDepth        if (depth > maxDepth) {            maxDepth = depth;            leftVal = root->val;        }    }        //Entry function    int findBottomLeftValue(TreeNode* root) {        int maxDepth = 0;        //Initialize leftVal with root's value to cover the edge case with single node        int leftVal = root->val;        findBottomLeftValue(root, maxDepth, leftVal, 0);        return leftVal;    }};

拓展：如果是返回最底层最右边节点的值呢？或者是计算数的最大深度？