Binary Tree Level Order Traversal II

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https://leetcode.com/problems/binary-tree-level-order-traversal-ii/#/solutions

这题跟层序遍历没有什么不同,只是把最后结果reverse 一下而已。不知道还有没有其他的套路?

可以分别用BFS 和DFS,后者要更简洁一点。

BFS:

/**
 * @param {TreeNode} root
 * @return {number[][]}
 */
var levelOrderBottom = function(root) {
    var queue = [[root]];
    var ret = [];
    while (queue.length !== 0) {
        var top = queue.shift();
        var tr = [];
        var tq = [];
        for (var i = 0; i < top.length; i++) {
            if (top[i]) {
                tr.push(top[i].val);
                tq.push(top[i].left, top[i].right);
            }
        }

        if (tq.length > 0) queue.push(tq);
        if (tr.length > 0) ret.push(tr);
    }
    return ret.reverse();
};

DFS:

var levelOrderBottom = function(root) {
    var ret = [];
    function iter(node, level, ret) {
        if (!node) return;
        ret[level] = ret[level] || [];
        ret[level].push(node.val);
        iter(node.left, level+1, ret);
        iter(node.right, level+1, ret);
    }

    iter(root, 0, ret);
    return ret.reverse();
}

 

写这题没花什么时间,但是写构造二叉树的代码和测试倒是挠头了一下T_T

通过层序数组构造二叉树,中序遍历和层序遍历

function TreeNode(val) {
    this.val = val;
    this.left = this.right = null;
}

function makeTree(arr) {
    function makeIter(i, limit) {
        if (!arr[i]) return null;
        if (i >= limit) return null;
        var node = new TreeNode(arr[i]);
        node.left = makeIter(2*i+1, limit);
        node.right = makeIter(2*i+2, limit);
        return node;
    }

    return makeIter(0, arr.length);
}

function middleOrderTraversal(node) {
    function iter(node, ret) {
        if (!node) {ret.push(null); return;}
        ret.push(node.val);
        iter(node.left, ret);
        iter(node.right, ret);
    }

    var ret = [];
    iter(node, ret);
    console.log(ret);
}

function levelOrderTraversal(node) {
    var queue = [node];
    var ret = [];
    while (queue.length !== 0) {
        var top = queue.shift();
        if (top !== null) {
            ret.push(top.val);
            queue.push(top.left, top.right);
        } else {
            ret.push(null);
        }
    }

    console.log(ret);
}

 

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