【算法15天：Day15】第六章二叉树 LeetCode 翻转二叉树（226）

题目二：

解法一：（递归）

js中可以解构赋值，交换很简单。

var invertTree = function(root) {
    if (root === null) return root
    let temp = root.right
    root.right = invertTree(root.left)
    root.left = invertTree(temp)
    return root
};
或者
var invertTree = function(root) {
    if (root === null) {
        return null;
    }
    const left = invertTree(root.left);
    const right = invertTree(root.right);
    root.left = right;
    root.right = left;
    return root;
};

// 前序遍历
var invertTree = function(root) {
    if(root === null) return null;
    [root.left,root.right] = [root.right,root.left];
    invertTree(root.left);
    invertTree(root.right);
    return root;
};
// 中序遍历
var invertTree = function(root) {
    if(root === null) return null;
    invertTree(root.left);
    [root.left,root.right] = [root.right,root.left];
    invertTree(root.right);
    return root;
};
// 后序遍历
var invertTree = function(root) {
    if(root === null) return null;
    invertTree(root.left);
    invertTree(root.right);
    [root.left,root.right] = [root.right,root.left];
    return root;
};

解法二：（层序遍历、广度优先遍历）

var invertTree = function(root) {
    const invertNode = function(root, left, right) {
        let temp = left
        left = right
        right = temp
        root.left = left
        root.right = right
    }

    let queue = []
    if(root === null) {
        return root
    }
    queue.push(root)
    while(queue.length) {
        let length = queue.length
        while (length--) {
            let node = queue.shift()
            invertNode(node, node.left, node.right)
            node.left && queue.push(node.left)
            node.right && queue.push(node.right)
        }
    }
    return root
};

解法三：（深度优先遍历）

var invertTree = function(root) {
    let stack = [root];
    while(stack.length > 0){
        let cur = stack.pop();
        if(cur === null) continue;
        [cur.left,cur.right] = [cur.right,cur.left];
        stack.push(cur.right);
        stack.push(cur.left);
    }
    return root;
};