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getCommonAncestor(el1, el2)复杂度 #5

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runshenzhu opened this issue Apr 10, 2015 · 0 comments

runshenzhu commented Apr 10, 2015

ml.js 中 function getCommonAncestor(el1, el2)函数
两个节点先求深度，深度大的向上走到深度小的节点同一深度。。。然后两个节点一起往上走，直到相遇。这样算法复杂度是否能小一点？
每次contains判断，(貌似)需要遍历整个子树，增加不必要的开销。

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