给你二叉树的根节点 root ,返回它节点值的 前序 遍历。
示例
示例 1:
输入:root = [1,null,2,3]
输出:[1,2,3]
示例 2:
输入:root = []
输出:[]
示例 3:
输入:root = [1]
输出:[1]
示例 4:
输入:root = [1,2]
输出:[1,2]
示例 5:
输入:root = [1,null,2]
输出:[1,2]
题解
建立一个栈,先把root节点放入
stack = [①] res = []
取出栈中的一个节点,记录值
stack = [] res = [1]
判断①是否有右子节点有则推入栈
stack = [②] res = [1]
判断①是否有左子节点有则推入栈
stack = [②] res = [1]
取出②,记录值
stack = [] res = [1,2]
判断②是否有右子节点有则推入栈
stack = [] res = [1,2]
判断②是否有左子节点有则推入栈
stack = [③] res = [1,2]
取出③,记录值
stack = [] res = [1,2,3]
③中没有子节点,且栈为空结束循环
/**
* Definition for a binary tree node.
* function TreeNode(val, left, right) {
* this.val = (val===undefined ? 0 : val)
* this.left = (left===undefined ? null : left)
* this.right = (right===undefined ? null : right)
* }
*/
/**
* @param {TreeNode} root
* @return {number[]}
*/
var preorderTraversal = function(root) {
if (!root) {
return []
}
const stack = [root]
const res = []
while(stack.length) {
const node = stack.pop()
res.push(node.val)
if (node.right) {
stack.push(node.right)
}
if (node.left) {
stack.push(node.left)
}
}
return res
};
来源:力扣(LeetCode)
链接:https://leetcode-cn.com/problems/binary-tree-preorder-traversal
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