【复习】LeetCode 二叉树DFS TypeScript

DFS（Depth First Traversal），深度优先遍历

前序遍历：根节点-左节点-右节点

中序遍历：左节点-根节点-右节点

后序遍历：左节点-右节点-根节点

前序遍历

TypeScript 复制代码

/**
 * Definition for a binary tree node.
 * class TreeNode {
 *     val: number
 *     left: TreeNode | null
 *     right: TreeNode | null
 *     constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
 *         this.val = (val===undefined ? 0 : val)
 *         this.left = (left===undefined ? null : left)
 *         this.right = (right===undefined ? null : right)
 *     }
 * }
 */

function preorderTraversal(root: TreeNode | null): number[] {
    const res:number[] = []
    const dfs = (root: TreeNode | null):void =>{
        if(!root) return
        res.push(root.val)
        dfs(root.left)
        dfs(root.right)
    }
    dfs(root)
    return res
};

中序遍历

TypeScript 复制代码

/**
 * Definition for a binary tree node.
 * class TreeNode {
 *     val: number
 *     left: TreeNode | null
 *     right: TreeNode | null
 *     constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
 *         this.val = (val===undefined ? 0 : val)
 *         this.left = (left===undefined ? null : left)
 *         this.right = (right===undefined ? null : right)
 *     }
 * }
 */

function inorderTraversal(root: TreeNode | null): number[] {
    const res:number[] = []
    function dfs(node: TreeNode | null):void{
        if(!node) return 
        dfs(node.left)
        res.push(node.val)
        dfs(node.right)
    }
    dfs(root)
    return res
};

后序遍历

TypeScript 复制代码

/**
 * Definition for a binary tree node.
 * class TreeNode {
 *     val: number
 *     left: TreeNode | null
 *     right: TreeNode | null
 *     constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
 *         this.val = (val===undefined ? 0 : val)
 *         this.left = (left===undefined ? null : left)
 *         this.right = (right===undefined ? null : right)
 *     }
 * }
 */

function postorderTraversal(root: TreeNode | null): number[] {
    const res:number[] =[]
    const dfs = (node: TreeNode | null):void =>{
        if(!node) return
        dfs(node.left)
        dfs(node.right)
        res.push(node.val)
    }
    dfs(root)
    return res
};

共勉