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LeetCode: Binary Tree Inorder Traversal
LeetCode: Binary Tree Inorder Traversal
Given a binary tree, return the inorder traversal of its nodes‘ values.
For example:
Given binary tree {1,#,2,3}
,
1 2 / 3
return [1,3,2]
.
Note: Recursive solution is trivial, could you do it iteratively?
地址:https://oj.leetcode.com/problems/binary-tree-inorder-traversal/
算法:要求用非递归来进行中序遍历。初始时,从根节点开始,一直往左走,并把每个节点入栈。在while循环内,取栈顶元素,出栈,访问该元素,若该元素有右孩子,往右走,在一直往左走到低并把每个节点进栈,如此循环知道栈为空。代码:
1 /** 2 * Definition for binary tree 3 * struct TreeNode { 4 * int val; 5 * TreeNode *left; 6 * TreeNode *right; 7 * TreeNode(int x) : val(x), left(NULL), right(NULL) {} 8 * }; 9 */10 class Solution {11 public:12 vector<int> inorderTraversal(TreeNode *root) {13 vector<int> result;14 if(!root) return result;15 TreeNode *p = root;16 stack<TreeNode*> stk;17 while(p){18 stk.push(p);19 p = p->left;20 }21 while(!stk.empty()){22 p = stk.top();23 stk.pop();24 result.push_back(p->val);25 p = p->right;26 while(p){27 stk.push(p);28 p = p->left;29 }30 }31 return result;32 }33 };
LeetCode: Binary Tree Inorder Traversal
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