每日一题（15）

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    vector<int> postorderTraversal(TreeNode *root) {
        vector<int> arr;
        stack<TreeNode*>tree;
        TreeNode *cur=root;
        TreeNode *last=NULL;

        while(cur || !tree.empty()){
            while(cur){
                tree.push(cur);
                cur=cur->left;
            }
            TreeNode *top=tree.top();
            ////右为空 或 右树根==last表示有树已经访问过了
            if(top->right==NULL || top->right==last){
                arr.push_back(tree.top()->val);
                last=top;
                tree.pop();
            }
            else
                cur=top->right;
        }
        return arr;
    }  
};

给定一个二叉树，找出它的最小深度。最小深度是从根节点到最近叶节点的最短路径上的节点数。

思路：
如果root为空，直接返回0。当节点左右子节点都在的时候取最小；当只存在一个子节点时，取最大；

class Solution {
public:
    int run(TreeNode *root) {
        if(root==NULL){
            return 0;
        }
        else if(root->left!=NULL&&root->right!=NULL){
            return 1+min(run(root->left), run(root->right));
        }
        else{
            return 1+max(run(root->left), run(root->right));
        }
    }
};

迭代方法2：

class Solution {
public:
    int run(TreeNode *root) {
        if (root == NULL) {
            return 0;
        }
        queue<TreeNode*> queue;
        queue.push(root);
        int depth = 1;
        while (!queue.empty()) {
            int num = queue.size();
            for (int i = 0; i < num; i++) {
                TreeNode *top = queue.front();
                queue.pop();
                if (top->left == NULL && top->right == NULL) {
                    return depth;
                }
                if (top->left != NULL) {
                    queue.push(top->left);
                }
                if (top->right != NULL) {
                    queue.push(top->right);
                }
            }
            depth++;
        }
        return depth;
    }
};