打卡-算法训练营-Day13 ｜二叉树的递归遍历；二叉树的迭代遍历；二叉树的层序遍历

leetcode链接

144.二叉树的前序遍历：leetcode.cn/problems/bi…

94.二叉树的中序遍历：leetcode.cn/problems/bi…

145.二叉树的后序遍历：leetcode.cn/problems/bi…

递归遍历

简单题，复习一下前、中、后序的遍历方式以及递归的写法

1. 前序遍历：根 -> 左 -> 右
2. 中序遍历：左 -> 根 -> 右
3. 后序遍历：左 -> 右 -> 根

/**
 * 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) {
    let result = [];

    // 前序遍历：根 -> 左 -> 右
    let recursion = (node) => {
        if (!node) return;
        result.push(node.val);
        if (node.left) {
            recursion(node.left);
        }
        if (node.right) {
            recursion(node.right);
        }
    }

    recursion(root);
    return result;
};

中、后序代码类似，修改result.push(node.val)的位置即可

迭代遍历

使用栈结构实现迭代遍历，前序和后序还好理解，中序不太好理解，一刷先知道下思路

前序遍历：

var preorderTraversal = function(root) {
    if (!root) return [];

    let result = [];
    let stack = [];
    stack.push(root);

    while(stack.length) {
        let node = stack.pop();
        result.push(node.val);
        if (node.right) {
            stack.push(node.left);
        }
        if (node.left) {
            stack.push(node.left);
        }
    }

    return result;
};

中序遍历：

var postorderTraversal = function(root) {
    if (!root) return [];

    let result = [];
    let stack = [];
    stack.push(root);

    while(stack.length) {
        let node = stack.pop();
        result.push(node.val);
        if (node.left) {
            stack.push(node.left);
        }
        if (node.right) {
            stack.push(node.right);
        }
    }

    return result.reverse();
};

层序遍历

var levelOrder = function(root) {
    let answer = []
    let queue = []

    if (root == null) {
        return answer
    }

    queue.push(root)

    while (queue.length !== 0) {
        const currentLevelSize = queue.length;
        answer.push([])
        for (let i = 1; i <= currentLevelSize; i++) {
            let node = queue.shift()
            answer[answer.length - 1].push(node.val)
            if (node.left) queue.push(node.left)
            if (node.right) queue.push(node.right)
        }
    }
    return answer
};