刷LeetCode刷到遍歷二叉樹的題目造寝，前序（題號144，中等）吭练，中序（題號94诫龙，中等），后序（題號145鲫咽，困難）签赃。遞歸比較簡單，但是題目建議可以用迭代分尸，三道題用簡單的遞歸也都可以通過锦聊，迭代的思路難一點。

JavaScript實現(xiàn)
前序-遞歸法：

/**
 * Definition for a binary tree node.
 * function TreeNode(val) {
 *     this.val = val;
 *     this.left = this.right = null;
 * }
 */
/**
 * @param {TreeNode} root
 * @return {number[]}
 */
// 前序遍歷箩绍，遞歸法
var preorderTraversal = function(root) {
    var res = [];
    res = getNode(root);
    return res;
};
var getNode = function(root){
    if(root === null){
        return [];
    }

    var arr = [];
    arr = [root.val].concat(getNode(root.left));
    arr = arr.concat(getNode(root.right));
    return arr;
};

前序-迭代法：

/**
 * Definition for a binary tree node.
 * function TreeNode(val) {
 *     this.val = val;
 *     this.left = this.right = null;
 * }
 */
/**
 * @param {TreeNode} root
 * @return {number[]}
 */
// 前序遍歷孔庭，迭代法
var preorderTraversal = function(root) {
    var cur = root;
    var res = [];
    var stack = [];
    while(cur || stack.length!==0){
        if(!cur){
            cur = stack.pop();
        }
        res.push(cur.val);
        if(cur.right){
            stack.push(cur.right);
        }
        cur = cur.left;
    }

    return res;
};

中序-遞歸法：

/**
 * Definition for a binary tree node.
 * function TreeNode(val) {
 *     this.val = val;
 *     this.left = this.right = null;
 * }
 */
/**
 * @param {TreeNode} root
 * @return {number[]}
 */
// 題目要求用迭代，先寫個遞歸看看
var inorderTraversal = function(root) {
    var res = [];
    res = getNode(root);
    return res;
};
var getNode = function(root){
    if(root === null){
        return [];
    }
    if(root.left===null && root.right===null){
        return [root.val];
    }
    var arr = [];
    // 左中右材蛛，中序遍歷
    arr = getNode(root.left).concat([root.val]);
    arr = arr.concat(getNode(root.right));
    return arr;
};

中序-迭代法：

/**
 * Definition for a binary tree node.
 * function TreeNode(val) {
 *     this.val = val;
 *     this.left = this.right = null;
 * }
 */
/**
 * @param {TreeNode} root
 * @return {number[]}
 */
// 迭代
var inorderTraversal = function(root) {
    if(root === null){
        return [];
    }
    var res = [];
    var stack = [];
    var cur = root;
    while(cur!==null || stack.length!==0){
        if(cur !== null){
            stack.push(cur);
            cur = cur.left;
        }else{
            cur = stack.pop();
            res.push(cur.val);
            cur = cur.right;
        }
    }
    return res;
};

后序-遞歸法：

/**
 * Definition for a binary tree node.
 * function TreeNode(val) {
 *     this.val = val;
 *     this.left = this.right = null;
 * }
 */
/**
 * @param {TreeNode} root
 * @return {number[]}
 */
var postorderTraversal = function(root) {
    var res = [];
    res = getNode(root);
    return res;
};
var getNode = function(root){
    if(root === null){
        return [];
    }
    if(root.left===null && root.right===null){
        return [root.val];
    }
    var arr = [];
    arr = getNode(root.left).concat(getNode(root.right));
    arr = arr.concat([root.val]);
    return arr;
};

后序-迭代法：

/**
 * Definition for a binary tree node.
 * function TreeNode(val) {
 *     this.val = val;
 *     this.left = this.right = null;
 * }
 */
/**
 * @param {TreeNode} root
 * @return {number[]}
 */
// 迭代
var postorderTraversal = function(root) {
    if(root === null){
        return [];
    }
    var res = [];
    var stack = [];
    var cur = root;
    while(cur || stack.length!==0){
        if(cur){
            stack.push(cur);
            cur = cur.left;
        }else{
            // 檢查還有沒有右邊
            if(stack[stack.length-1].right){
                cur = stack[stack.length-1].right;
                stack[stack.length-1].right = null;
            }else{
                cur = stack.pop();
                res.push(cur.val);
                cur = null;
            }
        }
    }
    return res;
};

上面遞歸的方法其實沒必要用兩個function的圆到，不過算了懶得改了怎抛。⊙康看得明白就行